get focused refocused on this we're going to do 7.4 so here we go the question on Section 7.4 that puts this question on your mind is how in the world or can we not where in the world how in the world or if in the world this is possible magic one you what I don't want it to be possible let's just stop let's just not do it uh and the idea is yes this is possible uh we can do integrals like this if we can do a couple things with them and I I'll talk about those couple things right now the section is entitled partial fractions and the name implies fractions now we're going to take this fraction and we're going to break it into parts and we're going to do integrals of now how we break up this fraction into Parts is very specific it depends on the number of cases uh the reason why this is possible is there's there's two rules in algebra with rational functions that you need to know so here's um here's how we'll call this the how how is based on two rules from algebra rule number one every polom in the world pols have U positive Powers they do not have square roots they have positive whole number Powers they're not like X to the3 they're not x to the 1/2 pols are things like this and things like this those are pols if you don't remember what those are those are pols every polinomial in the world can be factored as a product of linear and or irreducible quadratic factors that's a that's a law in well not a law that's a a a theorem approved theorem in algebra so every polinomial can be decomposed uh decomposed means broken up in different parts can be decomposed or can be factored into a product of linear and or irreducible quadratic irreducible quadratic vectors basically here's what this says if it's linear linear and or irreducible quadratics it means this ax + B and or ax^2 + BX + C every polom in the world can be factored as stuff that looks like this and or stuff that looks like this there are some quadratics that you can't Factor we know this correct hopefully we know this some quadratics have complex factors complex Roots which means you can't Factor them with rationals or whole numbers you can't do it but we can definitely reduce them into either linear and or quadratics this is our idea right now does that make sense so when we look at this you go hey look at that I don't know if I can factor that one don't really care but I can definitely definitely Factor this one it's a power three does that make sense I can factor that it's going to be linear and or quadratic factors problem number two or sorry rule number two rule number two is that every rational every rational function so basically where you have R ofx over Q ofx where R and X are both pols so for every rational where the degree of R is less than the degree of Q can be decomposed into partial fractions now there's one I I know I'm giving you a lot of theory right now and not a lot of practice don't worry we're going to get a lot of practice on this stuff um I want to give you one last little bit here look up here let's make sure that we understand a couple of these Concepts okay first first concept this is polinomial this is polinomial hence this is a rational are we all okay on the terminology the degree of this polinomial is two it's it's the largest power the degree of this polinomial is good if the degree of R ofx the top the main the numerator here if the degree here is less than the degree here you can decompose this into partial fractions so far so good mhm if the degree here is greater than or equal to the degree here then you do long division of pols first and what that does is that reduces the degree of the numerator to something that is less than this degree does that make sense so if the degree here is less than this you're good if the degree here is more than or equal to this you do long division first and then it reduces the degree of your numerator and then you do partial fractions those problems are kind of stinky we'll see those a little bit so little note no there's not a shortcut for degree r ofx syntheis no shortcut not synthetic division sure if it works if it satisfies the conditions but that is long division just quicker no I just quicker for degree RX greater than or equal to degree QX do long division first are you ready for the next little bit I know there's a lot of writing right now we're going to get to this example I promise but are you okay with this so far you kind of have to be okay these are two theorems that are are proved in mathematics every polinomial can be factored into a product of linear and or irreducible quadratic factors that's these and or these every rational where the degree of r is less than the degree of Q the numerator is less than the degree of the of the denominator you can reduce that into partial fractions if you don't have that you do long division which brings your power down your your degree down then you do partial fractions that's the whole deal there are several cases that we're going to discover happen with these type of fractions uh the first case is by far the easiest case and we'll talk about the easiest one first and we'll gradually make it a little bit more difficult and see what happens here so here's case one okay case one case one is where we have distinct linear factors distinct means unique it means you don't have multiple of the same factors otherwise you'd have that linear Factor squared does that make sense so distinct would be like you have x + 1 * x -1 * X +4 what you don't get you don't get x + 1 * x + 1 * x - 4 that would be non-distinct then you'd have x + 1 s we call that repeated linear factors are you all seeing the difference here so distinct means doesn't happen again happens one factor and then you have a different Factor so distinct linear factors distinct linear factors here's what we do in that case if we have distinct linear factors then what we can do is we can write r r x over Q ofx equals we break this up and we write a over ax A1 X Plus B1 and then we have another one b b over a 2x + B2 and we do this for every different linear Factor that we have notice this is linear you guys with me it's not quadratic doesn't have a power two this is linear this is linear so basically what we do is we Factor our denominator we see what factors we have if they're all distinct linear factors what we do is we make up a fraction for every single one of those distinct linear factors I know this might be new for you but are you okay with the idea here are you sure so we've got polom we Factor the denominator we look look at what type of factors we have if they're all distinct linear we write a fraction for every single one of those distinct linear factors until we get to the very last one that we put whatever we have um oh you know what usually we use a and b and c and so on and so on and so on so at the very end we sometimes some books do this A1 and A2 all the way to a sub this is just we use a and b and c and d but for to make our notation a little bit nicer we'd have the N actual uh constant where A and B and C are all constants do you want to see how to do it in practicality probably I want to end this I want to end with this example so hopefully we can do this so we leave our integral alone right now we don't we're not even worried about it we're not even doing any calculus right now it's all algebra so we're going to take the the uh integrant here and we're going to worry about the denominator right now and see if we can Factor it completely here's what you can and what you can't do you can't just Factor one little thing and assume you're done you've got to kind of show that it's completely factored you can't just Factor one little bit and go ah I'm pretty sure it's completely Factor you got to know okay it's got to be completely factored which means that we don't ever leave cubics ever uh we we often we sometimes leave quadratics but you better make sure that if you have a quadratic it's not factorable it's irreducible are are you still with me yeah okay so first step number one step don't worry about the numerator I don't care about that what I care about is factoring the denominator right now let's factor that how can you factor XB - x^2 - x pull the X out so if we do that we're going to get x * X 2 - x - 6 still okay so far of course you are it's basic algebra right all right now do we stop here you better make sure that this is not factorable if you're going to leave a quadratic is this factorable yeah it is3 pos2 so we'd have x * X and doesn't it does not matter the order not one bit - 3 x + 2 okay show Advantage be okay with that so far now here's the idea what we know then so bringing this down is that we have now 4x^2 - 4x + 6 over what this is this is the denominator it's x * x - 3 * X +2 we Factor the denominator completely no no saving factors till later no just ignoring them we factored completely this is factored completely quick hit not if you're okay with that one so far then you look at your factors how many factors do we have three three are they all unique if they weren't unique you would have combined them wouldn't you you'd have something squared or something cubed or whatever they're all unique in this case are they all linear do you see any power twos there no so here's what we do next it's not a hard thing all you got to do is say what this is going to be equal to how many factors do you have three you're going to have three fractions you're going to have X you're going to have x - 3 you're going to have x - 2 we make a fraction for every sorry plus two I know something wrong you can have a a fraction for each distinct linear factor that you have show F if you okay with that so far on the numerator these are all going to be constants this is the idea here so this is going to be a we don't call this a sub two we just call it B to make things easy and this is going to be C are we okay with that so far now the idea from here is that if you know these things are going to be equal if you can find a common denominator here then our numerators would be equal does that make sense if you can find a common denominator so that's what we work on so what we do is okay cool uh there's an easy way to do it if you were to find a common denominator for this notice that it is going to be this ultimately that is going to be our common denominator well this is only missing these two factors this is only missing these two factors this is only missing these two factors I call this the cover up method it's kind of nice all you got to do is go okay well cool uh cover up the denominator and just write the other two linear factors so if I were to find that common denominator I'm going to cover this up then we we'd have to multiply in order to find the common denominator we'd be multiplying this fraction by x - 3 x - 3 and x + 2x + 2 does that make sense to you we're not writing the denominator because of this if I ever get like a 3X overx -1 = uh 5 + 1 5 + x over x - one if I ever get that I know that these things are going to be equal If 3x = 5 + x does that make sense to you of course we'd have a domain issue with one but that's that's what we we ultimately solve this problem show should be okay with that so I'm not writing the denominators here because I know that if I get this common denominator denominators are gone anyway so I've got a * x - 3 and then x + 2 plus let's see if you can do the b b * x x cool because I'm going to cover this up I don't need another x - 3 I've already got that factor there for my common denominator I am going to however have an X and X Plus still okay let's do the C what's going to be there for the [Music] C okay now check it out what's this whole thing going to be it's going to be equal to whatever this denominator is if I have 3x x + one or x - one whatever I gave you before uh it doesn't really matter and I say this is equal to 5 + x/ x + 1 whenever your denominators are the same you can solve that problem by setting your numerator equal you can even do that with the proportion cross multiply you'll see that the X Plus 1's are gone does that make sense to you so I have I would have numerator equal to numerator if these guys are the same the idea here is that what you're trying to get is a common denom you can do this the long way you really could but you know what you get you would get a x - 3 x + 2 plus b x x + 2 plus c x X - 3 you know what this is going to be over it's going to be over x X - 3 x + 2 so basically this thing equals this thing do you see how the denominators are the same here that means the numerators get set equal to each other this will be equal to this because your denominators are exactly the same does that make sense like cross multiplication it is a it's a cross product that's right it's it's a I get it it's ultimately just a proportion that you're creating uh we are ignoring this step because we know it's going to happen we know that your denominator if you create a common denominator your denominators are going to be the same look you're keeping the exact same factors aren't you Therefore your common denominator here is this thing if we just multiply by the missing factors we inherently create a common denominator that we can ignore because it's going to be the same level size question are we going to have to about um for the denominator like x z or 3 or -2 later on yes but that's the same thing that we would have worried about here right that that's in your domain so yes yes and no uh we we don't create any I'll say it this way we don't create any additional problems with this one does that make sense to you because this already has those problems in it and when you factor it you just see the problems so we yeah we have to worry about them for sure if we have those we're going to have what's called improper integrals that's the next section we're going to cover okay so we get there ultimately but right now we're right now we're ignoring that a little bit I'll be honest with you we're ignoring it uh we will talk about that in the next section improper integrals where we go wait a minute what would happen if we integrate from uh let's see 0 to three Well we'd have a problem at zero we'd have a problem at three won't care about the negative2 it's not going to show up but we'd have a problem with those things we will deal with that uh but for right now I want you to see that we're not creating any additional issues do you see that M these are the same right same domain issues do you understand the concept of when I do this I create a common denominator and when I do a common denominator I have my numerators set equal to each other do you get that idea okay we're going to stop right there for right now we'll continue right here next time well we're going to continue so where we left off last time was how in the world do we do an integral like that where a substitution won't work where is a polinomial over polinomial hence it's rational and where the degree of our numerator is less than strictly less than the degree of our denominator where it's not I'll show you what to do in the next uh probably the next couple examples here so if this is the case what we know by some algebra theorems is that we can reduce any fraction such as this into partial fractions partial fractions work in a case-by casee basis we're either going to have all linear factors or we're going to have linear and irreducible qu quadratics or we might have repeated linear factors or repeated quadratic factors that's the idea with this now in our case when we Factor it which you have to do by the way uh you got to know that any polinomial can be decomposed into linear Andor irreducible quadratics so you'll never end with a power three it'll be either power one like here here here or power twos that you can't reduce like x^2 + 4 you can't factor that so we'd be stuck with it that that would be an irreducible quadratic are you guys with me okay so in our case we have one two three linear factors what that means is we're going to create three fractions one for each of our linear factors the way we do it we just write each of those linear factors on a denominator and constants a b c and so on and so on and so on until you exhaust all of your linear factors so far so good after that we basically find a common denominator we want to make this have the same denominator as this and I explained this hopefully very well last time if you do that the numerators themselves will be equal does that make sense so we would multiply this fraction by these two factors since they're missing that's why we get a and then x - 3 x + 2 we multiply this fraction by these two factors that's how we get an x and x + 2 this fraction by these two factors x X - 3 this is the numerator of this this this is the numerator of this and now because we have a common denominator those two things are equal show up hands if you okay with this so far all right now we get to continue and what in the world are we going to do with this uh well system equation we're going to have to we're going to have to distribute all this stuff and organize our Co organize our variables so if we do that I'm going to move quickly because this is of course well just some distribution so here we go we'll have our 4x^2 - 4x + 6 we get ax2 we get let's see + 2x - 3x - a x - 6 a I want to make sure that I'm right so can you please double check these as I'm going through just to make sure I got this stuff right too okay here we'll have plus bx2 plus 2 b x + C x^2 - 3 CX I'd like a double check if you got that same thing I'm double check my work too did you get the same thing I got okay now organize them what we're trying to do here is we're trying to match up X squ and x's and constants so for instance what I want to do is I want to look at this x s and I want to look at these x s but right now it's a little bit confusing as what's going on so we're going to factor this out so check it we're going to have do you see do you see the like terms I'm talking about we have a ax s bx2 CX2 if we factor out the x s then what this gives us is A + B + C X2 do you understand the idea hello yes no yes okay so these we've taken care of our X squs are gone then we look for any of our X's I see a ax a positive2 BX a -3 CX so we will have I always do a plus here we're just going to factor these numbers out the the coefficients are our numers so we'd have a we'd have + 2 b we'd have -3 c x so we're looking at this we're factoring out x x x therefore we have ative A+ 2 B minus 3 3 C show F feel okay with that one okay cool so those are now gone I always Circle them too that way I don't get lost what I'm doing I I definitely don't want to get lost and you can't miss a negative here or you can't screw up the coefficient if you do that your problem's smoked all right you're done so we got to be really careful with our algebra lastly well this is it there's only a minus 6 a now this is what's kind of nice about this check it out what we know is that 4x^2 - 4x + 6 equals this thing don't we well look if we know let me have your eyes up here this is kind of important if we know that 4x2 here's X2 right our X2 coefficients have to equal so if we have 4x2 and we have a + b + C X2 then what we know is this coefficient must equal this coefficient it has to equal that so we know that 4 equal A + B + C that has to be the case that's why we organize our X squ and our x's and our constants so that we can make this jump so we go hey here's an X squ here's an X squ this is four this has to be four that's the idea otherwise these things will not be equal because I have no other like terms quick head not if you're okay with that one okay let's do the next one here I have X and I have X the coefficient of this x is what4 therefore -4 equal a + 2 b - 3 C just match them up now the last one's nice I know that my constant is 6 I know that my constant is -6 a therefore six has to equal -6 a so we've grouped all we've basically factored out our our like terms from our like terms so we have a plus b plus C x^2 hey 4X S 4 must match up with a plus b plus C -4 must match up with a + 2 b - 3 C POS 6 must match up with - 6A should I feel okay with the idea now here's whatever algebra you know to solve these things for instance start easy how much is a once you know a is ne1 you get to use that fact over here to create a system of equations so if a is-1 then here we have -4 = pos1 you see why it's going to be positive 1ga 1 POS 1 + 2 b - 3 C or if I subtract the 1 -5 = 2 b - 3 C also here's our -1 here so 4 = -1 + B + C or if I add the one we'll end up getting this five ALS b + C okay so far I'm going to have to move up here I'm going to rewrite this a little bit so what I know is that a is1 what I also know is that I've Got 5 = b + C I know that -5 = -2 sorry positive 2B - 3 C okay so are you familiar with how to solve systems of equations yes all right so I have a to be and a not to be how to solve this that is the question that's right it's really good right okay anyway what do I do to solve this thing multip by yeah I'd multiply something to get some coefficient the same on one of my variables in our case we're going to multiply this by two therefore we get 10 = 2 B yeah you knew it either way uh and then five and we get 2 B minus 3 C if you do that you'll just have to subtract them if you multiply by -2 you would add them doesn't matter which way you do okay uh I'm going to choose to subtract them because I'm going to get positives are you with me on this you guys okay with that so far yeah so I'll subtract all this stuff if I subtract it 10 minus5 is 15 2 b - 2 B is 0 2 c - -3 C is 5 C 5 C are you with me on that one if you added it all you end up getting if you multiply by2 and add it you get -5 you get5 C are you still going to end up with three either way am I going too fast on algebra no okay I hope not so then C equals 3 okay I got a = 1 I've got c = 3 can you find out what B is okay how do we find out what B is yeah just plug it into whatever works for you this one's going to be the easiest so I know if a is1 that's how I found out my two systems I'm sorry two equations by using a a elimination method I've got Cal 3 that's great now that I know Cal 3 5 = b + 3 or b equals how much okay I want to make sure that we are all okay with this so far it's just some algebra but I know the algebra can be hard for for a lot of people I want to make sure you guys are all all right with it so figure out a common denominator that way your numerators are equal distribute organize so factor out your common terms here so in our our common factors in our case x^2 x and just a constant set them equal to the appropriate coefficient and then solve in whatever way works easiest for you here this was nice this is easy use that use the easy stuff to make the hard stuff easier okay now that we have a system of equations use any you can do substitution you can do elimination I don't care it's all going to work out exactly the same all you need to be able to do is find out what these letters are however many of them you have you need to know those letters feel okay with that one now here's what this means what this means is that our original integral this is the cool part our original integral because our degree here is less than our degree here it meant that I could do this because I did this and now I found out my coefficients check this out this is the awesome part now this integral becomes hey how much is our a1x plus how much is our B 2 2 x - 3 how much is our c 3 + 3 x + 2 DX if you can make that jump you understand how to do integration by partial fractions do you understand that jump right now I need to make sure that you actually do show hands if you do understand that jump the jump is use some algebra whole bunch of this is the hard part this is going to be easy okay uh if I didn't tell you this already you have learned every technique of integration that we are going to cover here uh now we're going to apply them to improper integrals so it's going be a little bit of a change we're doing this but this is all algebra so what this says is use some algebra to figure out that this is equal to this if you figure out your coefficients a b and c then this fraction is equal to -1x + 2x - 3 + 3 x + 2 hey look at that a was 1 right there b is 2 C is 3 therefore since this fraction equals this sum or partial fraction this integral equals this integral of partial fractions so far so good it's kind of neat right now the rest of the stuff is really easy really easy a lot of times not a lot of times all the time with your linear factor with your linear denominators you're going to get a whole bunch of LNS so we don't want to I don't i' never like to write more than I absolutely have to so when I do this I do all these in my head um because I'm I'm used to doing the substitution if it takes you some substitution off to the side do it but here's the deal how I do this integral I'd say hey look at this the integral of 1x is Ln X Ln Ln absolute remember absolute value so this integral the integral of one/ x is Ln absolute value of x since it's negative we just have a negative out front it's the negative is that constant it's a negative one so if we have a coefficient the coefficient just stays there quick head and aren you okay with that one now this one by rights you should be doing a substitution shouldn't you however if you did a substitution on this the derivative is one which means you're not going to be changing that coefficient whatsoever if this was like a 2x - 3 yeah you probably have to do a little something with that I would probably still do to my head um I do a little substitution my head I I would know how to do that I get a 1/2 but you'd have to take that into account but notice that if it's linear you can always do that quick substitution and it's still going to be an Ln the integral is still going to give you an Ln are you following so in our case if you had to do u = x - 3 D DX okay that's fine do up to the side that's okay I'm going to be a little bit quicker on that because I'm going to assume at this level that you understand how to do this right now okay so you could split this up up and do it independently of each other here we're going to have a plus what's going to be out front and then Ln absolute value x - 3 then another plus this is our coefficient 3 Ln integral x + two just please please be careful on this part let me have your eyes on the board here real quick be careful of this if you had a two right here or some other coefficient this is not the correct integral this is not the correct answer what would happen is you would do i' do I'd still do it in my head but I'd get okay dual 2 DX that would give me a 1/2 that's what that would be can you follow that one yeah if you got that then you're probably pretty good at this uh and you're fine but if you're not getting like w what's it coming from then take some time on your own and actually do the substitution off to the side you can't get these things wrong well you can I mean I don't want you to don't get these things wrong the constants and the coefficients are very important here so if fence you feel okay with with this one so do you understand that if I have linear factors I'm going to get a lot of LNS because it's going to be one over a power x to the power one I'm going to get a lot of Ln here now let me change it back this was not the case just want to show that to you this was not the case uh how do I W oh yep sorry this one too how do I wrap this up done that's the whole idea what's the hard part is it the integral no no hard part is this doing all this stuff show P feel okay with this one now would you like it if I showed you a shortcut that applies to a lot of these problems especially the linears you want to see that yes okay remember these numbers we can cheat we can cheat a lot so you have this written down correct are there any questions on this should be pretty straightforward I wanted to give this to you by the way uh sometimes you have to do this technique because you can't do the cheating way I'm about to give you but with linear factors especially cheat at it here's how you cheat at it it's not even cheating it's just being [Applause] smart if you want to be smart about this this is an equation correct M which means it should be able to give my coefficients if I plug the right numbers in so what you do is you leave it in terms of factors like this and this and this and this you leave it just like this and then you plug in numbers that make your factors go to zero plug in values that make your factors go to zero for instance when I look at this I go hey um and you have to show this this little part if I let xals 0 check your out if I let x equal 0 this yeah that's ne3 but how much is that zero this whole thing is gone does it make sense this would go to zero if I let x equal 0 this whole thing is gone does that make sense if I let x equals 0 this is a * -3 * pos2 because X is now zero you followed on the other side if x is equal to zero this is zero this is zero how much is that then we would get 6 = a * -3 * 2 let me go over one more time to make sure you're with me okay plug in values that make some of your factors go to zero you will have repeated factors okay so I'm thinking hey uh the easiest one to plug in if I plug in zero this is not zero this is not zero but this is 0 times anything is what folks this whole thing's gone 0 time anything is zero whole thing's gone this is also zero this is also Z so if I plug in Z 0 0 6 got it a no problem3 no problem two no problem 0 0 gone can you solve that yeah so quick so 6 = -6 a a equals 1 hey is that what we got yep brilliant now we got a couple more to do don't do anything more with this right now by the way um I always start with this always I always get as many as I can sometimes you'll still have to distribute but at least you got a couple that makes that makes some of this kind of irrelevant if you have a if you knew what B was you could find C very easy without having to do systems of equations okay so this is a very F much faster way of doing things if you can sometimes you can't but if you can yeah try plugging in some numbers that give you zero or on some of your your uh your big terms here so big terms make some factors equal zero give me another number that would work three would be a great number to plug in do you see why okay so if I plug in three I always put what I'm plugging in so x = 3 and then I put a little couple uh couple dots are here if I plug in three let's see can you figure out what I'm going to have on the left hand side if I plug in three notice I'm plugging in three not zero this doesn't go to six again you actually have to plug in the three what do you get 30 30 we double check on the 30 do you get 30 plug it in plug in three I want everyone to do this right now plug in three 30 triple check 30 yes uh a 30 person check 30 you get so if I plug in three I get 30 equals let's look what happens check it out if I plug in three please follow me up here this is how you you're going to get this stuff quickly okay not spend 30 minutes on just the setup plugin three what happens here this whole thing zero plug in three what happens here does it go to zero no this is B * 3 * 5 see where the three and the five are coming from three five plug in three here what happens this is why we plugged in three to make our Factor zero you multiply by Z it's gone see how easy that makes it we now we have 30 = 15 B how much is B hey we're two for two it's cool once you do the last on your own what number will you be plugging in if you do left one your own -2 yeah very good not two but -2 be careful on evaluating right here make sure you have your signs correct it's a big deal you have to get your signs correct here I know it's algebra but remember people take calculus to finally fail Algebra I want that to be you make sure your signs are right what' you get when you plugged in negative two over here you got 30 you get 30 positive 30 yes I plug in -2 what happens to this term what happens to this term what happens to this term you get a c you get a -2 and you get A5 see where the numbers came from plug in that2 well then we got 30 = POS 10 C or c = 3 did we end up getting the same numbers we got over here which way is quicker this way this way is obviously quicker does it work all the time no it doesn't work all the time especially when you have nonreducible quadratics because if you have non-reducible quadratics then you can't plug in a number that gives you zero like X2 + 4 you can't do that that factor is not going to be equal to zero ever therefore you can't use this technique with that you can use it for parts of it so like if you have a linear Factor man use it for the linear Factor Make That zero at least it gives you some of those numbers you follow me so fans feel okay with this part after that man you go through the same stuff so let's try this now I'm stick with this technique when I can because obviously it's a lot quicker let's do a different example are there any questions on this before I erase it all because we're going to need some space for this stuff you ready okay so new example let's talk about X to 4 - 3x^2 - 3x - 2 all over x 3r - x^2 - 2x DX tell me something about this problem that was different that is different from the last one that we did what's different the degrees the degrees are different remember what I told you about alra algebra said that if the degree here is less than the degree here then it can be written as partial fractions I said if uh is the degree here less than the degree here no no if it's listen if your degree of your numerator is larger than or equal to the degree here so if this was even a degree three we'd still have to do the following step what we're going to do is shoot guys I mean think about it if your degree here is bigger than or equal to the degree here that means you can divide it so we're going to divide div it by long division and then what that's going to do is reduce that degree so if you have like the last example where the degree of the numerator is less than the degree of the denominator start immediately Factor immediately and go for it if you have like this example the degree here is bigger than or equal to the degree here then you have to do long division and then you'll get to start your problem finding does that make sense to you so we're going to do long division I'm I don't know if you are that good at long division I'm not sure that you've seen it in a while um but here's how you do it we are going to go quickly through it you can refresh your memory if you want to on some of the videos I have for intermediate algebra that's where we cover it in u the appendices of that class so here we go I'm going need more space how we do long division of polinomial is we take our numerator and we divide by our denominator now when you're dividing your numerator when you're dividing your numerator please keep in mind you need at least a placeholder for every potential power of x that you have so if you do this look at the board please if you do this and just go right on down the line right here you're going to have an error you're going to have a mistake because what happens is when I divide X to 4 uh with my x to the 3 I'm going to end up getting something here I'm going to get an x to the 3 somewhere so the problem is 0 x yeah we're going to put a it's like a place value holder saying hey how many X cubes do you have zero have none of them put zero and then finish it off - 3x^2 - 3x - 2 so this is the same as this we got X 4th no X cubes - 3x2 - 3x - 2 and this is our denominator so far so good okay let's continue how we start this thing is basically just the first term divided by the first term so what do I get when I divide X 4th by X the 3r X and then we distribute we get X 4th check this out this is why we had to have a place value holder we get ne or - x 3r and we get min - 2 X 2 it's got to match up if it doesn't match up then you've made a mistake you forgot to place Val holder somewhere after that what do we do change signs yeah you can we're subtracting you either put parentheses here or you subtract and you change all the signs I like the parentheses to show what I'm doing X 4th minus x 4 is 0 0 x 3r minus X 3r you can change your sign and make that a plus if you want this will be positive X Cub this will be -3x plus sorry uh- 3x^2 + 2x^2 is X2 and then we bring down our next term yes I'm going fast but you should know at least a little bit how to do this are you all still with me so far yes okay and we start over how many times does X 3r go into X3 and it's positive 1 so we'll put plus one and then we distribute we get x 3r - x^2 - 2x again we are subtracting again we'll have parentheses you either change signs or do it in your head I do it in my head x3 - x3r is 0 uh x^2 + x^2 is 0 -3x plus 2X is and we bring down Rus 2 now we check again does x 3 go into x no then this is our remainder that's our remainder here's how you write your remainder what this says is that this fraction is the same thing as x + 1 this is our quotient here plus the remainder ofx - 2 over and when you write your remainder it's always as your remainder over your original denominator are you okay with that so far mhm just don't make the silly mistake of doing this please let me have your eyes on the board right now so you see this a lot of mistakes here are not calculus mistakes they're sign mistakes they're sign errors if you do this and you go oh look at that minus uh Mr Le's silly why don't you just put minus I don't put a minus there because this sign is wrong right now do you know that that's incorrect so what I'll do is I'll put a plus with my negative I'll put this on the numerator I'll put my plus sign there because you can always add but later so here's what you'd have to do to change that sign you'd actually have to factor out your negative here does that make sense if you factor out your negative this becomes x + 2 and that's that's something that we can deal with so I'm going to any questions on Long Division at all okay so this is what we get so then we know our integral is now x + 1 no problem then look at this plus and negative that's minus x + 2 x 3r - x^2 - 2x DX so we do long division what long division does long division is kind of nice with this look at this it pulls out some terms some terms here that are really easy to integrate we got an X can you do the integral of x can you all do the integral of x it's a piece of cake can you do the excuse me integral of one absolutely and it gives us a fraction where the numerator the degree of the numerator has been reduced and that's the idea of division so our remainder here is what counts our remainder is what makes this process a little doable for us we have this remainder ofx - 2 factor out the negative before you change it sign that is a must you have to do that if you have a negative there so feel okay with that if it wasn't if it was just a positive x - 2 that's fine leave it as plus you're good to go but if that's a negative sometime often times I'll factor that that negative out it makes it a little bit nicer to see can we move on so what are we going to do about this thing easy easy maybe split up your integral and see what's going on here so that's what we're going to do I'm going to take and make a little asteris here on this part I'm going to keep the negative here now don't lose track that you have a minus when you have a minus whatever we get listen please whatever we get is going to be in a bracket this minus we're going to treat like a negative 1 being multiplied onto that bracket it's going to change all of our signs are you following me on this one so you can think of this put a circle or put a big bracket right now that's what I typically do and we'll pull this off to the side and we'll work with it so here's our here's our piece right here at the same time maybe we can Factor this guy let's Factor this what factors out of XB - X2 - 2x if we got that we got x^2 - x - 2 can I go further yes remember you don't just Factor once and say I'm probably good uh you got to kind of prove that if you have a quadratic it's irreducible that you can't that you can't Factor it uh that would be done by doing a discriminant and finding out that's negative if you do a discriminant and it's negative that means you have a complex roote and that therefore it can't be factored so you'd prove that or You' at least go through it in your head now this can be factored what is it okay we've done a lot of work so far but I want to make sure that we're good so we start the problem off we look at it we go oh degree is not less than this one that means long division I've done long division I erased it I know we do long division this is our quotient of long division with our remainder of long division how you write your quotient is you always write quotient plus remainder if we have remainder over our first denominator this denominator then if it's a negative maybe factor out the negative it changes it to a minus so we get x +1 minus x+2 over our original denominator and now easy easy ah is a degree less than the the degree of the numerator less than the degree of denominator yes now we can go ahead and do this what I want you to do right now I want you to write out what this should be equal to U make your different fractions however many that you need and put the appropriate letters up top it's it's a small step but it's important step I want to see if you can do it how many fractions are we going to get however many different linear factors you have that is how many different fractions you end up getting what's going to be our first denominator please denominator X and then X and then X does it matter which order you have these in so if you factor this different well you can't Factor differently but if you have your factors out of order compared to this one no big deal you're going to do the same thing it's just your pieces of your integral will be a little switched around all right same idea no problem addition is commutative so we're good uh what's going to be on our numerators what's this going to be a b and c so fans feel okay with this so far we've done a lot of algebra huh have we done any calculus no we're not going to for a while either so we got long division check it we got this no problem partial fractions we have it all set up remember that when you have linear linear is power one when you have linear factors we make a fraction for every different factor and we put constants up top up top a b c and so on and so on and so on until we exhaust all of our distinct linear factors still okay so far now go ahead I want you to do this on your own create the common denominator use the coverup method I taught you last time so what I want to see next I don't need to see one fraction equal one fraction I want to see x + 2 equals and then just give me what this is equal to go for it so use the idea of finding a common denominator once you find a common denominator that means your numerators get to be set equal to each other if you've already done that try finding a b and c use the easy way if you can if I find a common denom what that is is that you end up multiplying each fraction by the missing factors in your particular denominator so here I have a factor of X I'm missing a factor of x - 2 I'm missing a factor of x +1 for B in order to create a common denominator I have the factor of x - 2 I have that already I'm missing the factor of x and x +1 so we have plus b * x and x + 1 just the missing factors of your LCD of your common denom lastly we got plus C I'm missing X and I'm missing x - 2 show you got that far on your own good that's fantastic that's a whole idea now can you find a b and c by plugging in some different numbers hello can you do that tell me the three numbers you'd probably plug in here zero for sure and then what two would be one and then the other one yeah plug those numbers in if you haven't done already go ahead and do that so when x equals z that's nice some of your factors are going to go some of your terms actually some of your terms are going to go to zero when x = 2 some of your terms are going to go to zero when x = -1 some of your terms are going to go to zero unfortunately man you got a this is nice to plug stuff into right a square over there that's that's beautiful I'm start working on this for the sake of time but I want to understand if you if you I want to know if you guys understand the idea here so we're checking for numbers that change some of your factors to zero if your factors change to zero some of your terms are going to go to zero so when I plug in zero start from the left 0 + 2 is 2 a okay if I plug in zero here I get -2 and I get 1 so this is a * -2 * 1 1 plus well that's a zero now I'm plugging in zero this goes a zero this goes a zero I'm good to go now let's plug in two if I plug in two don't forget you have to actually plug it in here 2 + 2 is 4 equals if I plug in two 2 - 2 is z this whole term is gone 2 - 2 is z this whole ter term is gone I get B * 2 * 3 did you get 2 * 3 B * 2 * 3 good lastly if I plug in 1 remember actually plug it in 1 + 2 is if I plug in1 here this is zero whole term gone this is zero whole term gone this is not zero I get C * one that's what I'm plug it in time3 should you made it that far good deal now we're almost done here so simplify a little bit 2 = -2 a therefore a equal how much 1 4 = 6 B therefore b equals how much is it okay to get a fraction yes don't worry about it 1 = 3 c c = 13 hey look at that those were the three constants I was looking for a b and c what that means is that our our original actually right here our original integral changes to this please watch very carefully on how I do this I want to make sure all our signs are correct you guys are okay to this point right did this make sense to you yes now that we have our A and C check it out what I know is that our integral is X don't forget about this stuff + one minus I put my big bracket so this should look identical x + 1 minus big bracket now inside that big bracket we just did a partial fraction of this thing what I know is that the partial fraction of this thing right here equals ax + BX - 2 plus CX + 1 do I have my a my B and my C then what this is equal to is -1 /x plus just do the look at just do the fraction like this 23 x - 2 don't worry about that +3 x + 1 DX a big bracket let me double check to make sure you guys are good with it uh do you see how this becomes these three fractions we found ab and C right so if we know this is equal to hey a plus b plus C over these relative denominators we got1 overx perfect we get 2/3 x - 2 got it we got 1/3 x + 1 not a problem what are we going to do next integrate no we're not going to integrate now we're going to make sure our signs are correct so what are we going to do now distribute that so x + 1 + 1X - 23 x - 2 - 1/3 x + 1 DX on the whole thing are you ready for the last little part yeah interesting yes interesting we can actually do it huh kind of weird but all it is is algebra it's same it's rational break it up into fractions not even a big deal now that you've broken up into fractions these are all easy to do what's the integral of um X I'm going to write that as 12 x s you okay with that what's the integral of one what's the integral everybody should know this integral of 1x L abute value cool we got this one we got this one we got L and absolute value of x we're going to have a minus sign here's how you deal with this look at uh look at the board here real quick you keep your constant just like we did in the previous example so this will be two can write it correctly I thought it 2/3 2/3 comes out front remember if this was an integral by itself look it if this was an integral by itself so integral DX you pull out the 2/3 would you you have 1X - 2 you do a little substitution u = x - 2 du = DX there's no constant pulling forward you're good to go then You' resubstitute back in for = xus 2 it's the same idea we've been doing so 2/3 comes out we'd have an Ln we' have an absolute value x - 2 CL absolute value it's really you're you learn some of the basic U substitution so that allows you to do this in your head um otherwise we spend forever doing this stuff um so so yeah if you need it do it uh but if you don't if you can do it like this that that's fine we've learned enough about this to be okay with that minus minus what3 uhhuh are you supposed to forget those absolute values no absolutely not uh no and then 12 x s x Ln absolute value x 2/3 forward another Ln 1/3s forward another Ln show P feel okay with what we done awesome all right we're on to our next example let's take a look at this integral now of course the first things we check for this just like any integral is does it fit anything we can do that's easy does it fit a substitution of course not does it fit integration by parts no it doesn't trig trig sub no nothing like that then we start thinking about okay think about this as a rational as a partial fraction is the degree here less than the degree here well that's great when it wasn't what do we have to do when it's not if we do our long division then it reduces the degree of the numerator and we go for partial fractions after that I believe our last example was like that one right okay so if we don't have uh something that we have to do long division with we jump straight into the partial fraction what's the first step in doing partial fractions fact factoring yeah not the numerator I don't care about that but denominator for sure so let's go ahead let's Factor the denominator can you factor x Cub + x^2 - x -1 yes I can can you factor it what do you do when you have four terms you group them try grouping them unless you see an obvious Factor like uh of course a GCF like X number or something yeah okay cool fact of that uh but if you don't see that try factoring by grouping for four terms if that doesn't work you have to find a root so you'd have to guess and check for a root or find some other method there are more advanced ways to factor uh I'm not going to teach him here that's an a higher class on how to factor Cubix and cortic and to prove that quintics are not factorable it takes a long time to prove that uh but you prove it for five and higher you can't generally Factor it for four and lower you can't there's formulas for it um but for for us to make it easy you factor a GCF first if you can like an X if it goes into all the terms if not try factoring by grouping if not then you have to come up with something that actually works like a number that gives you zero and then you can factor out x minus that number and that's the idea you can do that by division or synthetic division so anyhow for grouping we'd have our X cubed I'm going to erase this in a second so make sure you have this down - x - one grouping works like this you group the first two terms and you factor out the GCF you group the last two terms and you factor out the GCF notice when I factor out negative one it changes my sign and then you factor this out and we get x + 1 and we get x^2 minus one we okay with that so far yeah now is this good enough for us what's the problem with that okay so when we say Factor we mean we imply factored completely factoring means this factoring means linear and or irreducible quadratics so if I end right here with x +1 and x^2 -1 this one's great that's cool that's linear linear means power One X to the first Power you with me yeah this one this is quadratic that would be cool if it was plus if it's plus you're done do you see why yeah you can't Factor X Plus X2 + 1 but if it's x^2 - 1 that means you have to keep going so we'd do x + 1 and we do x + 1 x -1 so if F feel okay with that one now when you factor something we do want to make this in simplest form so what we're going to do I'm going to write this uh a little bit backwards I'm going to write x - one then I'm going to write x + 1 yeah listen I know it's basic algebra but I want to make sure you're okay with it are you okay by fact with factoring by grouping yeah do you understand that we are not going to stop right here with this thing we're going to factor as much as we possibly can till we get down to linear or irreducible quadratic show fans feel okay with that so far then I like to rewrite it as linear factors first everyone does this linear factors first and then powers of linear factors and then irreducible quadratics and then power and irreducible quadratics that's typically the way that we we write these so we've got linear and then linear to the second power you guys okay with this so far so I'm going to change this I'm going to say no no no no this is not good enough this is actually x - one and then x + 1 to the 2 power any questions on this before I erase it so we have Factor by grouping we continued to factor we grouped our factors that were the same so X+ one we now this is different you see before what we had was we had unique linear factors remember that and we set this up as a x -1 and then b x + 1 stuff like that well when we have these repeated linear factors that's actually case number two so case two is when we have repeated linear factors here's what we mean by repeated it means that you have more than one of them so for instance we have uh x -1 we had an x + 1 and we had another x + 1 that's why we had this X+1 squ this Factor was repeated hence the square if it was repeated G we have to the3 power you guys with me on this one so here's how you do it with any repeated linear factors what we end up doing is we write A1 over the first Factor plus A2 over that factor to whatever power we have and we continue until we run out of the repeated factors so for instance if this was to the fifth power what you'd end up doing would you you'd have A1 over whatever linear fraction you okay and you'd exhaust those powers of linear factors I'll show you in a minute how to do that if we had a 5th you'd have A1 over the factor a a different one so B you'd have B over the factor squared then C over the factor to the 3 then D over the factor to the 4th then e over the factor to the 5th you have every single different power of that factor are you with me on this one let me show you how it works here for us so in our case what we're going to end up getting is how many fractions do you think we're going to get here we're still going to get three that's right so we're still going to have one look at this two and again three do you understand why we have three fractions here yes we still have those three factors initially we still do it just one was repeated and so what we're going to have what do you think is going to be my first denominator x -1 x - one's going to be there you guys with me and then what we do is we say okay have we exhausted all the powers of x - one yeah there was just one of them we're good now we move on to the next one x + 1 X+ one's got to be here have we exhausted all the powers of x + 1 no so we go x + 1qu very good okay let's see if you're paying attention what if this was to the 4th power keep going two more i' have X+ 1 3r X+ 1 to 4 does that make sense to you so this one hey we're done with that one this one would be to the first then to the second then to the third and to four until you get all the powers up there until you exhaust all of them you with me on that one because we have these different factors if you had this you really have x + one and then another sorry x -1 then x + 1 * x + 1 * x + you'd have five factors there we show those powers to show those different possibilities of those factors now where did I stop what was that one two okay good now fortunately for us they're still linear right they're still linear so all we need on the numerators are A and B and C and then we do exactly the same thing we did before so now we have two cases we have the cases where we have distinct linear factors where we have x - one and it doesn't happen again we have X plus one and it doesn't happen again that was our first case and that case is pretty straightforward uh we just have every different Factor on the denominator a b c d e whatever until you you exhaust them you guys with me on that one then now we have the second case we've got the case where we might have some repeated linear factors for those linear factors that aren't repeated do the same thing exactly the same for those ones that are repeated have the first Power then have the second power this is the idea of partial fractions this is how it works you guys okay with it now can we make a common denominator still be smart about it when I'm making the common denominator remember the idea is make a common denominator then ignore the denominator so ultimately we're going to have 2X + 3x + 7 equal to whatever I get upon finding my common denominator here's the whole deal though listen listen and look you're going to have with repeated linear a lot of the same things aren't you use the largest power here so when I find my common denominator on this fraction of course I'm not multiplying by this guy I already got that guy but look over here I need to match this up I'm not missing x -1 I'm missing the x +1 sared don't take both of them just take the largest power does that make sense to you so I don't need this and this this one kind of counts for for this one as well this this is the only thing only Factor I'm missing you can see that on this previous fraction in order to find a common denominator with this I would just have to multiply by x + 1 to the thir or to the second second to the second so far so good it's just algebra but I want to make sure you're good with it plus so we covered up we take the largest power that is what's given us our common denom Now we move on to the B what's missing on this B fraction x - okay x - one is missing look you can look here if I cover that up what's missing here well x X - one is missing from there what else is missing from here squares the squares okay so do I multiply by X x + 1^ S would that be a good idea no no no don't do that that's way too much what should I multiply by you're only missing one factor of the X+ one are you guys getting that so this no so when I look here yeah I was missing x + one but more than that I'm missing X+1 s that's why we're multiplying by this I'm missing this whole Factor here with the B I've got one factor of X+1 already I'm only missing one more this look at this this would give us the x +1 2 and this would give us x -1 so I'm missing x -1 and one factor of x +1 from the C let's look at the C what am I missing from here to here what's missing that's it so I wouldn't take this one I've already got it it's like trading cards you go uh I got two Jose conos I got another Jose cono do you need it no I don't got two of them I don't need another Jose cono let you just want to make some money and sell cards but we already got those we don't we don't need extra one we have enough of our Jose conos that's a weird analogy wasn't it anyway question so because you already have the x + One S on the bottom you only need xus one for C that's correct basically we're looking at these fractions and some people teach it cover this up take the rest of it okay not this one but the largest power um I prefer you to think of we're finding a common denominator so look here look here what's missing between here and here x x + 1 to the second power is missing we need X+1 the second look here look here what's missing x -1 and one factor of x + one does that make sense look here look here what's missing that's it not X+ one you already have the X+1 squ honestly show pant feel okay with with that so far now did I teach you the the uh plug up that cheating way yes is the cheating way going to work for here no put one why not can you make this zero yeah of course you can can you make this zero yeah can you make this zero yeah can you make this zero then why would that work it works for linears works really well for linear all right it doesn't work for quadratics uh we'll we'll learn about that in a minute so why don't you go ahead and do that as I do on the board see if you can find the correct coefficients of a b and c so tell me the numbers that you are probably going to be plugging in here one and definitely negative 1 we try those right off the bat because negative 1's going to give us some zeros one's going to give us some zeros we'll be able to find at least two of these things for the last one I'll teach you how to do that in a minute so plugging in what did you already do them go for it I really want you to try it plug it in one if I plug in one 1 - one is z this is gone 1 - one is z this whole thing is gone one oh 1 + one that's two squ so all I know is that 12 = a * 2^ 2 did you guys get that as well 4 a I think a is three did you get a is three yeah awesome now do negative 1 do the one's first first so1 if I plug in NE 1 what I know is that on this side I get oh man what is that 2 - 3 + 7 how much did you get six okay so if I plug in negative 1 1 + 1 is 0 whole term gone 1 + 1 is 0 whole term gone 1 oh1 minus one how much is that2 so C * -2 what's c equal now here's an issue did it give you a b so I plugged in the numbers that give me zero I got a and c I didn't get B here's a little trick uh when you get down to here and like wait a second with these repeated uh linears this happens a lot okay because you're going to have stuff that looks like this where there's not a number there's no X there's no number you can actually plug in um oh you know what while I'm thinking of it as well so let me think of this right now if I had x to the 3 power like this if I had x to the 3 that would count as repeated linear this is not irreducible quadratic this repeated linear you'd have something over X plus something over X2 plus something over X 3r you have a d and e and f does that make sense to you so X the 3 is repeated linear X and then X2 and then X 3r okay just while it was on my mind wanted to clear that up I some of your problems are like that so let's say that we don't have a number that gives us zero for one whole term just pick another number to plug in something easy I would plug in probably zero just plug in zero you can plug in any number you want correct if I plug in zero yeah it's not going to get rid of anything but it's going to make it easier to do so i' plug in xal 0 check it if I plug in x equals 0 what happens on the left hand side s if I plug in xal 0 over here well all I'm going to get is a * 1^ 2 or a * 1 then I'm going to get B * let's plug in zero here if I plug in zero I'm going to get negative 1 I'm going to get positive one plus if I plug in zero here I'm going to get C * 1 ladies and gentlemen did you follow that just pick another easy number to plug in for these linear it's kind of nice if I plug in well not one I already dealt with that negative one I dealt with that those were the easy ones right those give us lots of zeros take another number you plug in anything you want to you plug in seven or 12 or -5 or one well you already did it or zero one in different case but plug in whatever you want and that way you get well an equation seven on the left hand side over here we get 7 = a - B- C hey this is why we did this first do you know how much a is yeah do you know how much B is no not b not B do you know how much C is yeah plug him in and find it so once you deal with a couple of these factors the rest of them become pretty darn easy as far as linear go oops so we've got 3 minus B and then plus three because we have minus a negative looks like we're going to get what Al together this would be 6 subtract the 6 we get 1 = b or b = 1 okay honestly that's the hard part finding out the A and the C at this point hey we're ready to split this thing up so it okay with this so far so what we know is that our original integral of 2x^2 + 3x + 7 over x 3r + x^2 - x - 1 DX is now equal to a brand new integral let's put this thing all together what we knew is that partial fractions said this can be broken up like this one two three fractions what it said was my denominators are x -1 that's going to be easy x + 1 that's going to be easy x +1 s well a little different but not such a big deal on the numerators it said that well provided you find a common denominator we did and can solve for these coefficients a b c we did then you know the coefficients of your fractions or basically the numerators of your fractions three don't get these out of out of order okay three from a and then negative one yeah negative one goes here you guys okay with that and then lastly it's three I'd probably clean this up a little bit we just want to make sure our numbers are correct so put them in the same order you did it okay so x -1 x + 1 x + 1 2 we double check x -1 x + 1 x + 2 we want a over x -1 so 3 X - 1 we want b/ x + 1 so -1 x + 1 we want c x +1 2 -3 x +1 2 so far so good maybe clean this up cuz it's kind of ugly looking so instead of plus negatives let's make this integral of 3 X -1 - 1 x + 1 - 3 x + 1^ 2 DX let's see if you remember how to do these remember how to do this integral that's not bad at all what's that going to be sure this is a you sub we do in our head the derivative of X+ 1 is simply one so DX = du it'll be deal so this would be a constant 3 Ln x + one in absolute value so far so good minus what very good so no problem no problem uh is this also an Ln is this also an Ln be careful this is not an Ln Ln is a quantity to the first Power this one maybe think of this as a different integral maybe pull that three out pull this x + one to the -2 if you were given this in calculus you'd be able to do it this is actually pretty easy uh I don't know if they teach you yeah they teach you substitution and calculus one don't they you'd be able to do this just fine don't forget the stuff you know don't make it harder than it is don't be throwing LNS all over the place when it's not an Ln this is hey that's a power that's a -2 so when we do this we' have three because of our constant we'd have x + 1 to the -1 because you add and then you divide by 1 because you divide by your new exponent what that would become is -3 over x +1 I want to verify that you guys are okay with this this would be -3 over x + one yes no you sure okay so notice how we took this piece away so I have minus that piece3 x + 1 clean it up just a little bit we'll have 3 Ln absolute value x -1 - Ln absolute value x + 1 plus 3x + 1 and finally let me know I'm done oh kind of I'm going to show you one more thing with this here's what I want to know are you guys okay with the algebra of getting our partial fractions yes no are you okay with integral of this guy and this one how about that one that should really be okay right cuz we're we're used to this bring up that exponent sure no problem x + 1 -2 uh do a use you do a little substitution but it's just x + one no big deal if this was like 2 x + 1 you'd have a 1/2 are you with me so here I mean no problem I know I'm doing it in my head I get it I'm not supposed to maybe but it's a that's straight substitution so we' have a u to the1 we have -3 over U then we' have3 x +1 you reset that right back in there since we're subtracting a negative we change to a plus add a plus c on there now one little thing I want to talk about occasionally you're going to get some problems that look different when you're doing these um repeated linear factors or even linear factors because what happens is you get a lot of LNS being added and subtracted and with coefficients out front now what we know from our chapter six work with linear oh sorry with exponentials and logarithms is that these really mean exponents this really means a division so occasionally you'll get stuff like this you say okay well this can translate to just to make it nicer looking Ln of x -1 to 3 over x + 1 in absolute value + 3 x + 1 + C your exponent this is how you would do this in order your exponent would move up it would become an exp going on x - one subtracting two Ln is a division problem notice the Q does not go to both it goes just to this first one your numerator are you okay with that one can't really do much with this one so you leave it but that's another way you could see the answer here we sure we're okay with that how you feel all right with that you guys ready to move on a little bit there any questions comments or concerns am I making this straightforward enough for you guys understand it okay well there's a couple of cases that we got to master so let's continue moving on most of these are going to take us at least one space of the board so if there's any questions I need to know now because I'm race most of it you got it all that nice work here we go oh my okay well that looks fun let's go ahead and do that one let's run through the whole process proc of solving integrals all right because on your test that I'm going to give you I'm not going to tell you what to do you're going to look at and go oh yeah okay I know what to do here so let's pretend this is a test situation you look at this problem and go go ahead and do it and you go uh what do you check for first the denominator I would check for substitution honestly that's what I would check for is this the derivative of this thing disregarding constants it's not but I check that first because if it is that's an easy problem that's really really fast you with me i' check it first after that you go okay well does it look like anything we've had before any like a uh integration by parts no it's not a product those are typically products um does it have any trigonometry in it any square roots or something we can turn into trigonometry and completing the square no it's a cubic partial fractions yeah looks like a partial fraction to me okay something we could use partial fractions for the first thing we check I think you guys said it first what do we check the degre yeah degree on the numerator has to be less than the degree on denominator or you do what with it yeah and that reduces the degree of the numerator for you so that's great so here we're good to go no long division necessary thankfully uh okay now that we've determined that we're going to do partial fractions what's your first step break it up break it factoring yeah we're going to factor it and we're going to factor by Aon said we're going to factor by grouping why don't you try Factor by grouping go for it I'm wonder if you get the same thing I did did you get 2x - one and then did you get x^2 - 4 plus oh good I was texting you you passed did you get a plus4 if you got let me have your eyes on the board here real quick you you guys okay with the grouping by the way you'll notice what I'm writing is the linear first you see that now if you had an x^2 minus 4 would you stop right here no if you have an X2 + 4 do you continue going no can you factor X2 + 4 no if you start factoring this as x - 2x + 2 your problem is going down the the poooo hole all right that's uh that's not good we can't start factoring like we don't know what we're doing here so this is your first case of an irreducible quadratic that's this next case right here case number three is when you have irreducible quadratics irreducible quadratics so we've factored by grouping it worked we had 2x - 1 no problem our remaining factor is x^2 + 4 this one okay what's what is this is this a quadratic or is this a linear linear you're automatically done there's no more factoring than that you're done is this a linear or quadratic quadratic can you reduce it more then it stays an irreducible quadratic for irreducible quadratics what we do we write the irreducible quadratic if we have powers we'd write it again and again so we write the first irreducible quadratic and then the second irreducible quadratic 2 two two and then the third one and so on until we exhaust all of our irreducible quadratics and on the numerators this hopefully this makes sense to you whatever Factor you have your numerator should be 1° lower than that so for linear one de lower than a linear is a constant 1° lower than a quadratic is a linear so for linears we'd have that's what we had here we had a we had B we had c one de lower does that make sense to you for our our linear for quadratics we're going to have linear ax plus b CX plus d and so on and so forth and so on and so forth this is the idea here so P feel okay with with that one now let's put that into practice with our problem here so notice how our cases are kind of they're building right we're building on what we already we already know so if you have a linear and you have an irreducible quadratic you're going to do two different things in the same problem how many fractions are we going to have here what do you think are there any factors to a power no okay so this is one factor this is another Factor are they repeated no no okay that's our next case Okay is when you actually repeat these so is this repeated no that's just one factor is this repeated that's just one factor you can't reduce it right so it's irreducible quadratic that means you're going have two fractions okay tell me folks what's going to be on my first denominator 2 is that linear or quadratic I know I've asked but is it linear quadratic okay what's on my next fraction is that linear or quadratic quadratic it's irreducible remember you got to make sure that you have simplified this or factored this as much as possible show hands feel okay with this so far it's the same idea you have a number of factors make a fraction for each factor that's it now on the numerators it's important you get this right get it correct for every linear we put a constant for every quadratic we put a linear one degree lower than what we have we do that for every single fraction typically this is why I write my linear first and then my quadratics so I don't lose any okay so I have to go back and forth so here what's going to be here that's right what do I want to do an ax plus b no it's linear I want a constant for linears how about this one do I put just a b BX plus C good what I want is a linear for a quadratic so constants for our linear that's what we've been doing here folks that's that's what this is then a linear for my quadratic notice how the degree is one lower this is X to the 1st this is X to the 0 this is X to the second this is X to the 1st it's 1° lower that's the whole idea should P be okay with that one then you're good to go that's it same stuff after this so why don't you guys go ahead and tell me what I would do to get a common denominator on my first fraction what's the missing Factor here remember the way I do it I look here and I look here what's missing from here to here so I have a and x^2 + 4 plus I look here and I look here what's missing from here to here 2 so the way this looks is BX + C remember it's the whole numerator time I'm sorry what what's missing there 2 so again for the last time hopefully when we find a common denominator we multiply by the missing factors on both the numerator denominator we do the missing factors are x^2 + 4 we have this one we're missing x^2 + 4 so that's why we got our a * X2 + 4 here we're missing the 2x - one we already got this we're missing the 2x - one that goes here and here as soon as we get that common denominator we ignore it we just set the numerators equal to each other so this numerator equals this new numerator this would be the numerator you follow just keep in mind it's a * this one this x to x - 4 and then it's this whole BX + C * the 2x -1 so feel okay with that now this is what I was talking about as far as sometimes you can cheat and sometimes you can't right now it's going to be pretty difficult to cheat at this thing and plug in a number that's going to work and the reason why it's going to be a little difficult to cheat is because if we plug in stuff like oh I don't know a number to make this Zero we' have to come up with 1/2 1/2 is going to be difficult to plug in here and difficult to plug in here and difficult plug in over there okay you can do you can try it you can't plug in things like uh like zero if you want so zero would give you -1 * * C and it would give you uh 4 a so and it give you equals -21 but it's not going to eliminate everything all right so sometimes one of the easier ways to go is just to distribute and actually see what happens so that's what I want you to do right here um if you want to plug in some numbers and sometimes you get lucky do it that's what I would try every time if there's an easy number to plug in man do it but there's not sometimes you're left with actually Distributing combining your terms and going forward so go ahead and do that now distribute combine your terms I'm going to to do it while you guys are doing it [Applause] make sure I got that a 2x- BX plus 2 CX did you get the same thing yeah collect your terms I see a couple X squs so I'm going to have A+ 2B X2 collect some x's again how I do this I always put a plus makes it easier for our signs when we Factor we'll have b + 2 c x are you guys seeing what I'm talking about there we put our plus and then we collect our signs inside of our parentheses B and 2 C lastly we have plus 4 A minus c as our constant we got those ones did did you make it that far on your own fantastic what's nice about this is when we collect our terms we have x s hey we match it up over here we've got X we match it up over here with our coefficients we've got a constant we match up with our constant so you tell me hopefully you can figure this out if I put a + 2 B how much is that equal to one yeah where are you getting that one from this is the coefficient of x s isn't it mhm this is the coefficient of x s isn't it it's got to equal so 1 equal a + 2 let's do the next one um if I put B+ 2 cative b + 2 C how much does b + 2 C have to equal here since this is the coefficient of x and this is the coefficient of x you're right now if I put 4 a - c how much ises 4 a - c half equal the reason why we use pluses here it keeps our signs correct uh that way you just look here and here and here set equal to the coefficients themselves so this has got to be -21 sure feel okay with that one so far okay now you can solve this in a variety of ways uh the best way I think I found to solve this um I would probably do a substitution I'd probably set said this one I'm going to do a lot of math quickly my head CU I don't want to spend a whole lot of time teaching you algebra okay so what I would do is I would add the C subtract add the 21 I would get C = 4 a + 21 does that make sense I would do a substitution here I'd get b + 2 * 4 a + 21 my equals 1 the idea being if I can get A's and B's I can combine it with this A's and B's and create a system of equations and that will let me solve it algebraically so do our substitution no problem we got B cool two no problem 4 a + 21 that's our c = 1 let's do a little distribution so b + 8 a + 42 = 1 are you still with me folks okay now I'm going to get my A's and B's on one side I'm going to get my con the other side because that's how this one looks I'll probably write my 8 a first cuz that's how this one looks my - B = -43 you still okay with that this is what we have now A + 2 b = 1 8 a minus B = 43 you guys should know how to do systems of equations in one of two ways either you do a substitution which would work here just fine or you do elimination which is what I'm going to do here and that's going to work just fine so if I did elimination I'd multiply this by two I multiply by two so if I multiply by two we get a + 2 b = 1 we get 16 a - 2 B = -86 have I've done that right have I done that right yes then we what do we do in this case we' add we' get 17 a this is gone equals 85 man this is kind of nice now let's solve it how much is a85 1785 over 17 what is it five five five I'm just telling what you give me what is it okay does it matter doesn't matter right I mean it matters it doesn't matter from here can you solve the rest of them yes piece of cake if a is5 solve for all Sol for whatever you want uh you can solve for C by this one by this one actually would be nice you could solve for b by this one that's what I would do first so if I plug the5 here I'd be adding that dividing by 2 B is how much sure -5 + 2 b = 1 therefore B = three so I get this one and I got this one now working over here I do this for our C I plug in ne5 there hey look at that 4 * 5 is -2020 + 21 is 1 that was relatively painless not so bad right not as easy as plugging numbers in uh but not horribly bad hey now that we have our coefficients can we rewrite our integral I know we've done a lot of algebra okay but I'm going to come right back to here so our integral x^2 - x - 21 over 2x 3r - x^2 + 8x - 4 dear goodness me equals an integral of two fractions now notice we have three three coefficients we have three constants but we don't have three fractions we have two fractions the first fraction is over what please the first fraction is over what the second fraction is over DX on the numerator of our first fraction it says that we're supposed to have only an a what goes here on the numerator of a second fraction we supposed to have a BX plus c 3x+ 1 beautiful we knew that this fraction can be broken up we we just showed it we know that a is5 hey5 2x - 1 we know that b is 3 we just showed it we know C is 1 plug those numbers in we get 3x + 1 x^2 + 4 so be okay with that one now I've got to erase this stuff do you have any questions so far on this you sure okay this very straightforward very straightforward this is going to be an Ln do you guys see it linear Ln the only difference is this is be -5 Ln absolute value 2x -1 it's going to be5 Ln absolute value 2x -1 over two do you see where the over two is going to come from it's a little U sub Right Here If U equal 2x - 1 du = 2 DX or du/ 2 = DX the over two is from right there does that make sense to you I do those in my head if you need to show your work you show your work don't get those wrong now I'm going to show that to you in a little bit this one some of you guys are struggling with Concepts like this man how in the crap am I going to do what am I going to do there think back to what fractions are made up of fractions can always be split up always now let's think it through okay think it through what would you do here basic UB or just do in your head look at this one what would you do here another that's a UB what's the derivative of an x s I don't care about the constant it's going to be an X do you guys see what I'm talking about how about that one let's think about that one okay we're going to do this over here do you guys want to finish this problem today yes okay we're it's going to take us about 3 minutes uh did you have questions on that too late okay let's just run down there I'm going to split these up because we're going this is crazy but we're going to use three different techniques on each of these three integrals so to show this accurately in mathematics you don't want to just do this all as one okay you have two different substitutions here are you following me so you're going to actually three you're going to split this up into three different pieces so integral of the first one plus integral second one plus integral the third one you could have pulled out the5 as well this integral is fairly straightforward r u = 2x -1 R du = 2dx our du/ 2 = DX therefore when you do your integral we get -5 Hales 5 Hales Ln absolute value of 2x - one and absolute value I need to show if I feel okay with that one this would be5 over U that would be5 Ln absolute value U you plug back in for your U and the DU over two gives us our five Hales so -5 Hales over U Ln U okay with that one no we're going quick but if you need refresher on that go back and look at that section online or on the video on the line okay next one don't let these things Escape you like this if you're stuck you go man I can't do that one try splitting it up then it makes a really easy integral doable for a hard integral doable for you makes it easier do a basic U sub here U = X2 + 4 d = 2 DX D over 2 = x DX there it is right there that means that we have three integral 1 over U du over 2 are we okay with that one yeah okay well that means that we're going to have three Hales Ln absolute value U so this will be plus use purple here to show that same thing plus 3es absolute Val sorry Ln absolute value of what goes in there and abute value I want to make sure you guys are golden on this stuff you guys okay with the first one five got a little substitution that's where the the over two comes from here basic substitution all basic du = 2dx d over2 = x DX this is Du over2 therefore we're going to have a three Hales that's the over two we're going to have integral of 1 U so we're having three Hales Ln absolute value U plug in back for the u u is x^2 + 4 we got our integral am I explaining this well enough for you guys okay next up you can actually do this two ways I'm going to show you a a very cool idea here that you're going to be using one idea is you understand that I could make this one over I could force this to factor as X over 2^ 2ar + 1 * 4 that would be 1/4 the integral of 1 /x +/ 2^ 2 + 1 and then you realize that's tan inverse and you go oh look at that that's tan inverse x 2 12 because you have a little use up there you can do it that way if you didn't catch that don't worry about so much I'll show you a different way hope that's right the way we'll see my constant might be up there I do it really fast here's another option this almost almost looks like a tri sub do you see it what's missing square root ignore it ignore the missing square root if you were to write it as a square root that would be the hypotenuse missing the square root did you get that you can still do that's a that's a tangent you can still do the two missing sides do you follow that if you do the two missing sides this would give you the X this would give you the two you can still do a substitution tan thet = X2 so who cares if it doesn't have the square root you can still do this no problem this is X this is 2 therefore this is square root X2 + 4 okay then X = tan th * 2 DX would equal 2^ 2 thet D Theta our integral would equal 1 over 2 tan Thea 2 + 4 our DX would equal 2^ 2 D thet 2^ 2 thet D Theta I know I just went really fast on this are you are you following that one you sure the triangle is the hardest part right then we do hey uh tan the X2 no problem x = 2 tan Theta DX = 2 c 2 D Theta make our substitution 1/ 2 tan thet 2 + 4 DX is 2 D that means that our integral becomes 4 tan^ 2 + 4 2^ S Theta D Theta Factor the four out if I factor the four notice it's even easier there's no square root here if we Factor the four we get 2^ 2 thet D Theta over 4 tan^ 2 + 1 do we follow that yeah hey look at that do you see 1/2 told you it was a 1/2 that's because of those constants hey2 integral hey how much is this so we're going to have secant s Theta over SEC s Theta that's equal to 12 integral oh my gosh this is square thet right this is square Theta right this is 1 12 square of what D can you integrate one so this is 12 Theta oh goodness gracious here's the whole banana is that even a term is that a thing you just made it a thing it's a thing whatever if tan Theta = x 2 this is if and only if Theta = tan oh my goodness this is where all your inverse integrals came from boah you just it then this is the same thing as 12 T = X2 correct = tan inverse of x 2 so 12 tan inverse x 2 I believe I had that on the board just a little while ago yeah so if you know how to factor that let's see you could do it with just uh your inverse trig function as far as your integral let go so one last little piece we're going to add this up here don't forget this is our first part we have plus 12 10 inverse X over 2 plus C now my question is did You Understand it yes somewh so this one Ln this one substitution then Ln this one you can do it two ways you can draw the triangle this still works don't even have to worry about the square root of okay just make the triangle accordingly or you could have made this factor out the four make it Factor the four you'd get an X2 2 + 1 and then a 1/4 outside do a little substitution for x over2 and you have an inverse uh tangent and that's the idea should okay with this one okay so before we get on to our last thankfully our last example of partial fractions uh I covered an example that had this little piece in it and I want to go over this a little bit more slowly uh on how you there's two ways to actually do this how you can go about doing this problem in one of two different ways now the first way the first way is an idea that we get from trigonometric substitution it's well you know what that looks almost like a triangle almost like one one hypotenuse of a right triangle you can actually do that so you could make a right triangle out of this and the way you can think about it is well if this were to have a square root around it like that this would be the hypotenuse is that does that make sense to you so it would be like this if that were the hypotenuse well the other two Dimensions here are the sides are the legs well we would have x and x and what two X and two where would the X go uh this one yeah because what what do we want to do here is make up our tangent relationship so we'd have X we'd have 2 and we'd have tan Theta = x/ 2 or x = 2 tan Theta and then of course we take our derivative we have DX = 2^ 2 Theta D Theta you guys okay with that so far mhm if you really think about it you can do this lots of times let's let's consider this one more time if you had just 1/x^2 + 4 someone made a comment earlier well maybe we can manipulate this a little bit so it's easier to see check it out if I have X do you understand that it's almost the same thing it's not the same thing it's almost the same thing as uh the square root of x squared mhm that's this is almost true uh it's not true because we'd Square it so our negative X's don't really work that way it's really the um it's really an absolute value type of idea but we can use it for here especially since we have both positive we don't have to worry about that there's no way we can get a negative out of this anyway so it's going to be positive no matter what does that make sense to you so whether X is positive or negative we get a positive out of it we would have to worry about that scenario so in our case we' say all right well if you want to make this a little bit more understandable about why this can fit a right triangle here it is why this can fit a right triangle is that we could make this the square < TK of x^2 + 4 squar now for those of you who didn't see it before do you guys see what I'm talking about that is that I mean that now that that is a hypotenuse right there it just now we have a squared do you follow me you can do that no problem we'd have our X we'd have our two we'd have hey that's my hypot now either way you go about doing this you're going to get the same thing because when we plug this in when we do our actual substitution well we've got one I'll leave it this way just so you see what I'm talking about we'd have one we'd have something being squared you with me inside here we'd have a square root we would have 2 tan Theta 2 + 4 and then we'd have our DX becomes 2^ 2 D thet you guys get me so far now look what happens with the square root and the square we didn't have to do this me personally I would have gone straight to my triangle from right here I wouldn't I wouldn't have needed that because I'm used to it but if you need this you go okay well this is the same thing as something with a square root squared yeah that's true that will allow you to see the triangle a little bit more easily let you set up this this relationship on our triangle let you find tan th = X2 no big deal but now watch it what happens when you square a square root take it away it goes right back to what you had does that make sense so you really didn't need to do it but you can do it if it helps you see it so in either case what we end up getting and I know that I did this earlier in class but uh we'll we'll do it again we get two secant s thet D Theta over what happens here is we're going to get I'm going to do a lot of math in my head right now just so you hopefully you're with me here uh the square and the square root are gone you follow I'd pull out the four I take a square root of four I'm going to get two I'd have a square root I'm sorry we wouldn't have a square root anymore the square Root's gone I made a mistake uh this is why I don't like to do the square root because it's it's gone now so maybe I can write this and you guys can follow me from here uh we' pull out the four and we'd have sec s Theta shoot I'm going too fast for my own good aren't I there we go that's gone we pull the four up of course we Square the two we Square the tangent pull the four up after this we'd have a 1/2 we'd have a secant squ Theta this becomes see 2 D this is 12 the integral of again I'm going too fast I'm doing the integrals in my head 1 D Theta don't even need the one what's the integral of I I went fast are you guys okay with getting down that far what's the integral of one yeah in this case is Theta so 12 Theta can I leave it in terms of theta no but here it is uh if tan thet = X 2 this happens if and only IFA = tan inverse of x 2 do you guys buy of that one tan X2 hey tan inverse of X2 so what we get out of this substitution is that our integral is 12 tan inverse x 2 + C this would be the long way of doing things but it is a way show fansy feel okay with the a way of do it you could do a trig sub just by looking here you can make it fit more of the trig sub idea or you could do this here's way number two way number two is to try to fit this to an integration table uh to try to fit this into integration table just the way it is and the way it is would be well if I can fit this to U ^2 + 1 du the integral of 1 u^2 + 1 du is tan inverse of U plus C of course well here's another idea then instead of going through the whole trig some idea well maybe we just think of instead of 1x^2 + 4 DX we force this thing we force this thing to factor out four here so we could say okay well if I'm trying to make this a one I've got to factor out a four you get me then I'd have 4 and i' end up getting in inside my bracket x^2 over 4 + one DX do you guys all the algebra on that if you don't just distribute right now and you'll see that this goes to X2 and this goes to + 4 you follow then what I need to do to fit this for I need to make this as something squared so x^2 4 no no no I'm going to pull the 1/4 out I would have 1 over X over 2^ 2 + 1 does that make sense to you so instead of X2 over no no let's think of this something squared let's do X over 2 2 you'll notice that 1/4 is outside out front this becomes X over 2 2 that's still x^2 over 4 and then the + one so far so good now you do just a basic U substitution basic so the u = x/ 2 du = 12 DX 2 du = DX and what we end up getting is 1/4 the integral 1 over what's the X over2 become U yeah u^2 + 1 and then we'd have 2D so instead of DX we get our 2D tell me next thing I'm going to do pull out the two pull out the two so I'd get 12 1/4 and two the integral of 1 u^2 + 1 d u hey do I know what the integral of U S 1 U ^2 + 1 is tan inverse U Tan inverse of U so this becomes 12 tan inverse of U how much is u plus c ends up being exactly the same thing so if you ever see and you are going to get this a lot uh listen you're going to get this a lot one of these things where you have 1x^2 + 4 or + 9 or + 1 if you see that how many ways can you go about doing it and either way is is fine U if you want the quick way and you want to do just a little substitution do this if you like the triangle method do this do do I care I don't care so as long as you can do one of these ways i' I'd say do both of them sometimes this is going to be nicer because if I do this like that you can still do it here you can't if I do this this is not this no don't do that so then you have to do something like along these lines so feel okay with this one okay now that was a little bit slower so but I just want to give you a heads up on what you can and can't do with this problem did that make did explain it well enough for you guys to understand it okay can we go on to our last example yes last one let's do it I'm glad you said yes I was going to anyway but now that I have your approval uh better by the way were there any questions on this uh too late we've done this stuff earlier in chapter 6 right uh we did this stuff earlier in Chapter 7 uh as far as the the trig sub um we also did yeah we've done we done all sorts of stuff in chapter 7 so somewhere along the lines we did this yes you guys got plenty of paper today right paper I hope you have a notebook full this is going to take a few pages of paper for you start a new page wait maybe start a new page I I'm going to need at least two of these board for this so heads up this going to take at least a gig or two of video at least probably probably two gigs this example two gig it's this it's this much see this and this one that's comp to don't worry you're going to know how to do all of it it's not hard it just takes a lot ofs yeah a lot of steps no long division because I'm a nice guy today see I like long okay so uh let's take a look at our example here let's see what exactly is going on remember that we have we have had three cases so far haven't we yes our first case was the easiest one you get a lot of LNS it's just it's it's linear linear factors the denominator is all linear factors like x + 1 2x - 3 those are linear then next case after that case two was you have linear but then you might have repeated linear and remember building up on the repeated linear you do for every Power until you reach the highest power uh so a over whatever B over whatever C over whatever you build that up now when we got to quadratics irreducible quadratics we no longer had just a constant numerator we had one it's did I tell you this it's always 1 de lower than what your your factor is so for linear we have constants for quadratics we have linears that's what our numerators are well this last case is not just Linear by the way can you tell me what the linear factor is here whatever that is turn it off now uh can you tell me what the linear factor is yeah that's linear we're going to have a overx right now you with me so we don't just neglect our previous knowledge of hey what do you do with linear you're you're kind of building on them so you can have more than one case in each example so linear sure now do we have any irreducible quadratics up here yeah that's an irreducible can you factor that no that's irreducible but we also have it repeated to a power and this is our case 4 case four is repeated quadratics so we had linear then we had repeated linears then we had IR reducible quadratics now we have repeated irreducible quadratics so this is our case four you forgot the DX I wasn't done yet so now that we're ready to go on our Temple uh let's see what what happens when we do this now before we go before we go and do this I want to give you an example of that's where're definitely not going to go through this whole thing all right I want to see if you understand how to build your partial fractions from this so little side step okay let's say that I had not even an integral just this thing right here if I gave you that would you know how to write the partial fractions to it this has every case that I can imagine here so I think it's kind of nice for us to go through this just so you see what's actually going on are you you guys with me on that one yeah first thing you check for is can you factor your denom any more can I factor this anymore this Factor not distribute vector vector no Vector no so these are all non-factorable do you believe me yes okay so they're completely factored now when we go through this you need to be good at identifying what linear factors are what repeated linear factors are what irreducible quadratics are and what uh repeated irreducible quadratics are the same method we use for repeated linear is the same thing we're going to do for repeated irreducible quadratics same basic idea so when we go through this oh yeah you know what the next thing we talked about you got to check the degree so how in the world do you check the degree here what's the degree of the numerator what's the degree of the denominator three lot more than a lot more than that a lot more because degree is after you've distributed everything right so let's start here you you'd get an X to 4th somewhere here wouldn't you and you get an x to the 6 then you get another x s so you have x to the 8 x to the 9th basically you have an X something X you have an X to 9th somewhere up here do you follow so naturally our degree here is less than the degree on a numerator so degree of denominator is did I say that backwards degree of numerator is less than degree of denominator and that's what we're looking for thank goodness because if it wasn't what would we have to do here comine it you have to distribute it and then do long division with it and then re the factoring would be crazy again after that but anyway this would be the idea should be okay with the the idea now let's see if you're good at telling me how many fractions we're going to get don't say that loud I want you to count them up how many fractions you're going to get okay don't say that loud everyone do this how many fractions are we going to get have you counted letters or fractions fractions how many fractions we're going to get let's see do you remember what to do with with our our factors here let's see what's the first one going to be and then I'm going to have another one with x- one oh so with repeated remember this when they're by themselves no problem when they're repeated you put the first Power you put the first Power correctly it's okay put the first Power then you put x- until you build up to the greatest power here so here we'd have X we have X - 1 we have X - 1 s that's what repeated linear was do you remember that let's just work on the all the fractions right now then we'll fill out the numerators what's next X you go to your next one so our repeated our quadratics so x^2 + 1 that one's not repeated so I'm good to go after that what x2+ x + 1 the same thing we did with our repeated linear so here we had x - one then we had X - 1 2 you take every Power up to the largest power so here x no problem X - 1 s well you got to have an X - 1 then you got to have an X - 1 2 x + x^2 + 1 no problem that's not two a power so we just have one of those factors here this is repeated so I have x^2 + x + 1 but then I'm going to have another one cuz it's repeated you have to go to power that you have and X2 + x + 1 s 1 2 3 4 5 6 those people said six you're right on the money if you didn't say six do you understand why we get six now you sure okay so we got this hey repeated you're going to have two of these that's what it says two one two of these so 1 2 3 4 5 six just count the powers up that should be how many fractions you actually get here so far so good okay especially if I would have some like this a lot of people screw this up a lot of people do this and they do that is that the appropriate thing to do you'd have an X and then you'd have an X squ so if you count the powers 1 2 3 4 five six seven this would give you those seven fractions that you should have on this example you okay with that one now let's go ahead I'm going to erase that again let's figure out what we should have what's the first letter we would start with it's linear next one good how about C this would be C yes that's still C because it's repeated linear every time we get a quadratic we have a power a degree less so power one power one this is repeat linear counts towards linear so we have a c we have DX plus e we have quadratic h a b c d e f g h h h fantastic that be the idea this is the idea show feel okay with this one this is why we're not doing this because this is crazy but you could if you really wanted to you could find your common denominator and you could figure out your a b c d e f g hi I Elemental P whatever you get out of there and you can do your partial fractions there's an easier way there's no easier way this is it this is the partial fraction reduction for this our decomposition as we we say now fortunately for us this one should be substantially easier than this one we have a lot less fractions to deal with what I want to know is does this actually makes sense to you guys so we count our powers make sure you have that appropriate number of fractions make sure that you don't do silly things like put a BX plus C here don't do silly things like for forget to do the X Plus don't forget the X's it's always got to be one less degree than whatever power that is this counts as linear you don't need a CX plus uh D here but for every subsequent one where we have our irreducible quadratics you definitely do need that any last minute question before we get on to our actual example okay the rest of this thing is going to take us a while so I'm going to erase this because I need the room but let's go back to this one so if we're looking at this then what we know is that XB - 2x^2 + 3x + 2 over x * x^2 + 1^ SAR can be reduced into some partial fractions or decomposing some partial fractions question is the degree of the numerator less than the degree of the denominator yes yeah this counts as X 4th here's X 5th that's more than x 3r that's great so we don't need to do long division or anything like that this is nice because that would be very hard to do with this problem we don't want to we don't want to do that we don't have to redistribute refactor okay can you tell me actually you know what why don't you do it why don't you write out the fractions that you're supposed to have here I think you guys are all at that level go forward I know there's at least one how many in total are we going to have three power one one two three yep you're right can you tell me what the denominators will be tell me what the denominators will be first what is it second denominator perfect last one fantastic that's exactly right show fans you get the denominator perfect good so repeated quadratics work just like repeated linears you just put the factors until you reach the largest power so here we have x to the 1 no big deal we have x^2 + 1 to the second well we need our first power we need our second power now let's fill the numerators out make sure we got those right what's the first thing we're going to write here is this going to have X's up here it's quadratic it's going to be BX plus C this one it's also quadratic did you get that yes does it make sense to get that we have linear should just be in a one power less or 1 de less quadratic should be 1 de less exus quadratic should be 1 de less still have an X joance if you okay with with this one are you sure you positive this is Big Time right if you don't get this are you going to get your problem right no no that's an issue okay we got to make sure this is this is perfect now moving on what's the next thing we do after we figure out how our rational expression is going to be decomposed in partial fractions what are we going to do find common denom let's find our common denominator why don't you go ahead and do it see if you can do it I'll do it up here in just a second so I'm only give you a little bit of time did you get it back there so remember what we do we look here we look here we multiply by the missing factors just like any other time in the world you would ever find a common denominator just like in algebra it is algebra this is all algebra okay I know I've said it about 50 times so far but your idea is to find a denominator between here and here if you can do that then if the denominators are the same your numerators must also be equal therefore when we multiply by the missing factors we get that common denominator automatically that's why we don't write these denominators here so we're going to have numerator equal to our new numerator um what I do I look here and I look here at each individual fraction I just multiply by what's missing so I know I'm going to have a and the thing that's missing between here and here is my X2 + 1 squ you guys follow me on that one between here and here well be careful I know some of you guys hate parentheses I don't know why you hate them they're friendly when you have BX plus C you got to have parenthesis there you will be Distributing so a * well I'm missing this whole x^2 + 1 2 got it BX plus C okay I'm missing what am I missing between here and here what's missing good I am missing X for sure and then one I like how you said that one factor of x^2 + 1 let's make sure I don't have the X so I got that now I'm missing oh I'm missing the square so I need one more factor that X2 + 1 if you multiply this times this it should give you this if you multiply this times this it should give you this that's how this stuff works so you're just multiply by the missing factors you guys okay with that so far and lastly we've got a well we have a DX plus e again in parenthesis so here we were missing this guy here we were missing X and one factor this guy here we're missing what so it's just got to be DX plus e * X show P feel okay with that perfect now I think I mentioned this earlier when you have these quadratic factors it's not as easy as when you have the linears with the linear was real nice because you can plug in numbers a lot of the times and eliminate entire terms like High sections of this you can't really do that all that well here there's there's one number I'd probably plug in if I wanted to just really quickly I'd plug in probably zero um if you plug in zero this whole thing's gone this whole thing's gone because X's would be zero here this would be 1 1 s is a if I plug in zero I get two right off the bat I know that a is two does that make sense to you so I probably do that first so if I do xals 0 I know that two so 0 0 0 here's two here's 0 + 1 is 1 1 2 is 1 a * 1 is a a = 2 all this stuff would go to zero 0 * everything would be 0 0 * everything would be zero so right now I know a is two so f beate with that logic if you didn't do that right away it's going to be very apparent to you later anyway so it's really not that big of a deal what do we do with the rest of this junk you distribute all of it no here's here's the problem if you if you're like well wait a minute why can't we just plug in okay tell you what uh give me a number here that you can plug in the makx is zero I'll give you a million dollars if you don't use complex dang it I yay I win yeah we're not talking about complex analysis are we class is horrible yet yet that's in upper that's later okay so no can you plug anything to make this zero no no can you plug in anything to make this I don't know because you didn't know what D and are so no the only thing we could plug in to help us here was Zero we're done with that so you're left with distribute you can distribute it distribute It Go For It distribute it and hurry because we're run out of time so I'm going to leave this off to the side here okay I know a is two that that's actually going to help us a lot that's good oh my goodness wow uh this will be fun you get to double check my word too so sometimes I make mistakes just like you guys I hope I don't you like distribute all everything like all of it you got to use you got to use P what about just the first one are we good x 4 + 2x^2 + 1 then a did you get the first thing like this too okay I know it's going to be a plus so let's see did you get that for the second one same thing yeah did you I want make sure I'm right too so don't leave me hang in here if you got something different and I did it wrong you let me know that's what I got because of the X I got that too man okay this one should be easy okay all right good got it did we get that that's all distributed you of course might take a little bit more time and do this and make sure that you're actually right because it makes no sense to do this and get it wrong otherwise your problem your smoke you spend 30 more minutes on this and get the wrong answer that sucks so make sure this is absolutely correct what do you do now combine kind of combine like terms you factor out the the like terms here so we look for our like terms and we do X Cub - 2x^2 + 3x + 2 no problem and then we collect them so I like to Circle them I know that that's kind of what well why are we circling them when we're in calculus 2 but I don't like to forget any I don't like to lose them so I Circle my like terms here like this one and this one um please remember that you're not factoring out the constants with your variable you're literally factoring out just your variable so if you had a two here and a two here you leave it with the A and the B because that's going to be a system of equations all I'm looking to have you guys do is do this a plus b do all the constants a and b r constants here and then factor out the X for does that make sense to you and we keep on going so I'm going to cross those out I don't want to ever make the mistake of writing it twice or forgetting one of them after that I'm going to look for any of the cubes do you see any cubes trust me on this you're going to love it when there's only one of them seriously you'll see why in a second I wrote C I thought c i i i don't know why it's a three man if I write a c when I think of three I get poed C3 P whatever come on that was good okay so I'm such a dor So fourth's Gone Cube's gone I get a 2B or not 2B C3PO I all I need is R2 and a D2 and we're we're good I'll start using RS next time you just need to make one with that many fraction no I don't want to do that okay our squares so with our squares I've got notice what I'm going to do I'm not pulling out the two I'm going to leave the two with the a so a 2 a and then a b and then a d shouldn't it be minus two I don't think so it's a positive oh I'm sorry I'm sorry I don't see any negatives up here so we're we're kind of fortunate as far I negative I'm sorry my mistake yeah something on that doesn't it anyway squ C plus e so we got our fourths we got our Cub we got our squares we got a 2 A plus b plus d and then X2 now we're going to look for our X's I've got that c and my e so plus C plus e x that's nice and then lastly we have our don't forget that a that plus it at the very end oh goodness okay what I want is show up pant if you're okay on the algebra of it is it going to take you some time but most of this is kind of algebra when we get down to the trick it goes very fast and that's nice all you're going to know how to do all the trig and all the and all the integrals no problem there will be trig in a lot of this by the way trick Subs uh or there's going to be inverse trick function like a tan inverse or something because when we get down to this guy and this guy you're going to be splitting that up you'll have constants over x^2 + 1 you're going to get that so be prepared for it that's why we covered all the previous sections now a lot of these are going to be nice and easy here's what I can show you do you know that your coefficients have to match up so for instance when I'm looking at this I'm looking here my only constant is a you got it my only constant over here is therefore a must equal we already knew that but do you see how this really didn't help us all that much it gives it to just right here anyway so let's list them all out over here I know that a is two so we got our a let's keep on going what's another easy one that you see I would look for C I look for all the individual ones look at the C I've got CX to the 3 you with me CX to the 3r look at my X the 3 uh term over there what's the coefficient of x the 3 therefore if I have X the 3 here and this is my only X the 3 how much does c have to equal the coefficients must be equal so coefficient here is C coefficient here is one therefore C equals 1 this was kind of nice right not that hard to do no no series systems of equations it looks like it's going to be ridiculous ends up being really easy you guys okay with that one then I start looking for stuff that's going to be easier after I come up these these basic ones so a got two c I got one if I see c and e well now I can figure out e really easily can you tell me what C plus e has to equal C plus e let's see if you could get this C plus e has to equal what sure C plus e x here's my X coefficient is 3 coefficient is C plus e therefore C plus e has to equal 3 okay if C + e equal 3 can you tell me how much E equals good C is 1 so now we're starting to use some of these basic substitutions C is 1 + e = 3 therefore e = two so I got a is two I got C is one I've got e is two can you figure out the rest of them from here yeah easily let's look at uh oh let's look at this one some of you guys aren't used to this this idea I've got a plus b x 4th correct wait a minute crap has to Beal to Z how many X the 4S do I have zero none so if I have a plus b x 4th and I don't even have an X 4th a plus b equals zero coefficients have to match the coefficient of x 4th would be 0 so 0 = A + B how much is a again two so 2 = b b = -2 hey we're almost done one last one I left the longest one for last not first because I don't I wouldn't know how to do it right I left this one for last so I'm doing easy to hard how much is 2 A + B + d equal to why don't you tell me 2 a perfect so X2 X2 coefficient is -2 coefficient here is this whole thing now we're just plugging in the numbers that we already know I know that a is 2 so 2 * 2 plus how much is B how much is D don't know yet so we're plugging in what we know we knew a was 2 2 * 2 we knew B was -2 so plus -2 then we just work it out simple algebra 4 - 2 gives you two how much is D have we found them all a b CDE e we got a B CDE e hey we got them all now let's go ahead and let's write out what these fractions are going to be so can you go ahead and do that for me please so write out what this integral changes into you should have the same number of fractions your denominator should be the same your numerator should be whatever coefficients you had whether it's a constant or whether it's a coefficient on X what's our first fraction be 2 next [Music] fraction oh okay so we want to make sure our signs are correct huh all right that's probably good idea so B is -2 so we're going to have -2X our c is 1 so a a was easy it's two -2x + 1 got it and our last one let's do DX plus C What's d so it's going to be -4x plus two plus two okay quick your hands feel okay with that one cool now we got to kind of move so I got to erase this you know I'll do one more step and I'll erase all this stuff whenever you get these quadratics or linear quadratics unless it works with a substitution which is not going to happen a lot okay because of that constant is not going to happen um so what we're going to to do here because if it did you'd have an X here uh you're not going to do that otherwise it would have been reducible so it's not going to happen what you do here split the fractions you got to split them up so our one two three fractions is going to change to this 1 2 3 4 five fractions you follow do not change your denominators ever so we have 2X here's what I'm going to do please watch carefully I'm going to have notice how I didn't bring down the minus because I have changeed it sign I don't want to do that I'm going to have - 2x over x^2 + 1 so I can leave this plus and have + 1 over x^2 + 1 I'm going to leave this minus and do - 4x over x^2 + 1 2 and then I'm going to have + 2 x^2 + 1 SAR DX so be okay with that one are you sure so we're splitting it up we got no problem split it -2X X2 + 1 no big deal uh we got plus 1x2 + 1 we get - 4x X2 + 1 2 + 2 notice how my denominators don't change I'm not changing that I can't split those okay but the num writers yeah we can now I'm going to do this kind of an interesting way I'm going to rewrite this whole thing I'm going to write this on the top of my paper and do each one right under it you guys with me so I'm going to have a lot of space here but I want to make sure you're seeing that I'm doing each of these integral rules separately you follow okay so keep this in mind I'm going to erase everything else so any other questions before we uh continue some of these are going to go very quickly some of these are going to take a little more time there's nothing I haven't taught you how to do so what I'm probably going to do is show you how to set the integral up and then let you go for it I'll probably write it very fast up here with answer is uh but you you guys at this point should know how to do all of these things so let's see by the way can I can I pull out some constants here and you guys be okay with it yeah so this changes to 2 * the integral of 1x you got it yeah DX I'm going to split up everything this is going to be minus 2times integral you can even leave the two because of our substitution we're about to do X x^2 + 1 DX we're going to have + 1 integral of 1 over X2 + 1 DX we'll have -4 you know what I need more space for this one oh that should be that one + 1^ 2 DX and then lastly we're going to have a + 2 * the integral of 1 x^2 + 1 sared DX oh my goodness that's a lot let's make sure we're okay with it we pull the two out we got 1X DX we pulled the -2 or the minus 2 out we separate our integrals we got xx2 + 1 DX this one was great just integral of 1x2 + 1 DX minus we pulled the four out we got xx2 + 1ty S DX pull the two out we got 1x2 + 1^2 do ex your pant be okay with that one oh my goodness now some of these are going to be very easy I'm going to go through quickly other ones I'll show you specifically this one and this one are the ones you're worried about a little bit the rest of them are going to be piece of cake so let's look at our cake can you do this integral 2 Ln absolute value of x no problem that's an easy one because we have a definition of integral of 1x DX is absolute Ln absolute value of x what would you all use for this just do a use sub if you got X so it's basically either going to be Ln a u sub or it's going to be a tree that's it so with this stuff we're going to do well u u = x^2 + 1 d = 2x DX du/ 2 = x DX so when we go for that we have integral of notice we have X DX this becomes our D over 2 so 1 / U du over 2 two keep in mind we have a minus two out front twos are gone we have minus Ln absolute value x^2 + 1 look I did that one really fast but I want to make sure you guys are okay with it are you okay with it you sure so this was you no problem basic substitution X DX is Du right over two right here two with the 1/2 gone integral of 1 U is Ln absolute value U plug the U Back In I know I'm going fast but you guys should be able to follow that or at least do it on your own you okay with that one yeah okay Moving On by the way are the absolute values pertinent here no you would write them and then you can't eliminate them if you want to because X2 + 1 can't be negative now this one this is an interesting one this one you either draw a triangle for or you understand that the integral of 1x2 + 1 is tangent inverse of X and that's very basic this right here is tan inverse of x no substitution no nothing that's literally in remember doing your integrals the first thing you check for is check for your integration table that's right there I'm check for it that's tan inverse you'd waste a whole lot of time if you were to draw your triangle it's kind of fun I like it uh but it takes a long time to actually do that right okay next one what would you do here same exact thing it be a Uub so U would = x^2 + 1 du would = 2x DX du/ 2 = x DX here's our X DX right here so this would be our -4 we'd have an integral we'd have 1 over U2 D over 2 here's the big thing I want you to notice check it out notice how this was our u in here correct so we have 1 over u^2 do you guys see where the U S com from is this going to be an Ln no this right here is our -4 sorry uh what should should be up there I mean two -2 this would be U to the -2 du we bring that up this is not always an Ln okay this one's not so -2 u 1 over1 or we get 2 over U if I've done that right hopefully I did that right so what that means look at what I did here okay I took the sign with it I took my minus 4 I changed it to A4 we did the whole thing it changed to positive two so I'm going to have a plus 2 over U what's our U X+ one that's the idea should can be located with with that one okay now where are we at now last one last one last one last one is not in your basic integration table we talked about this earlier the opening of our class this is one where you'd have to draw your triangle for so you can think of it think of this as your y as y hypotenuse x² + one I going do the whole class like that X2 + 1 and we got this uh leg over here x and this is if we got that thing then we're going to be using a tangent I'm not really going to do that that'll be fun though huh you guys get me on the triangle yeah that's got to be hypotenuse right if it is yeah we're kind of putting that square root up there we know we can we just change this to an X 4th and then that would be a square root does that make sense to you no you show that if you want to I'm going to kind of s side step that idea think just think about it think if this is part of a triangle I put a square root around it it's a hypotenuse then we have X and one because I'm trying to make tangent here it's either tangent or secant or sign not going to cover again we already did equals x well look at that DX thenal D okay so far can we finish this example up yes okay we're very very close what that means is that our integral here becomes two integral of 1 over tan^ 2 th + 1 2ar * SEC 2 Theta D Theta so this is one okay cool this was tan tangent 2 + 1 whole thing squar DX is see 2 D th show feel okay with that so far now what do we do so we're going to simplify yeah simple this is going to be D Theta I'm going to do this quickly this right here is SEC squ correct yes to the second power is to 4th still right so far that means that this is two 1^ 2 D th that becomes 2 cosine 2 D Theta see where the cosine squar is coming from yeah hey do we know how to do cosine Square [Music] 2 1 + cosine 2et D these are we get I'm just going to do the integral right here this integral therefore Isa plus sin 2 / 2 we keep on going remember that with your triangles I know you've seen this before you can't have a 2 Theta you'd have to make this that 2 sin Theta cosine Theta so we geta plus 2 sin Theta cine Theta over 2 twos are gone now we get to change everything back again so I'm going to do that over here once I get this if Theta sorry Tan = x Thena = tan inverse X got that one can you do do sign sign is opposite so that would be X over < TK X2 + 1 time oh cosine look at cosine cosine is 1 overun x^2 + 1 so from here this is all trig stuff it should go fast for you guys you should know it this is hey 12 1 plus cosine 2 2 th no big deal twos are gone we have integral of 1 is Theta integral of cosine 2 is sin 2 2 that's a basic little use of sin 2 no good 2 sin thet cosine thet so far so good awesome well then we have twos are gone Theta is tan inverse of x no big deal sin Theta cosine Theta multiply s cosine if you multiply your square roots are going to be gone this becomes tan inverse x + x over Square no more square root x^2 + 1 you guys see where the X over X2 + 1 come from you're multiplying here so then we're going to do plus tan inverse x + x over < TK x^2 + 1 lastly the only thing we do is combine some like terms here these two things you can combine oh yeah you know what you're right I'm sorry combine some like terms so we have 2 Ln absolute value x - Ln absolute value x^2 + 1 + 2 tan inverse of x + x + 2 over x^2 + 1 like term like term like term like term done