in this video we're going to talk about how to find the antiderivative of a function so what is the antiderivative well as the name implies it is the opposite of differentiation we know the derivative of f of x is f prime of x so the anti-derivative or the integral of f prime of x is f of x so f of x is the antiderivative of f prime of x whereas f prime of x is the derivative of f of x now the integral or the antiderivative of regular f x is capital f of x so capital f is the antiderivative of lowercase f and lowercase f is the antiderivative of f prime of x so hopefully that gives you a good idea of what the anti-derivative is so let's say if we have the second derivative of a function and we want to find the first derivative and we can go we can keep going forward in this direction i just want to give you an overview of these different symbols and functions so going to the right we're looking for the anti derivative going to the left we're looking for the derivative the process of finding the antiderivative is basically integration the process of finding the derivative is differentiation so hopefully that summarizes everything that you need to know about this conceptually now we said that capital f is the integral of f of x dx now you you get capital f plus c you always get the a constant of integration but we're not going to worry about that here what i'd like to do right now is distinguish what is an indefinite integral versus a definite integral so this here is what is known as an indefinite integral an indefinite integral gives you a function typically in terms of x but it could be in terms of some other variable like y or z or something like that a definite integral doesn't give you a function but it gives you a number so this is a definite integral the result of a definite integral is always a number like 20 negative 8 45 or something and you can easily identify or distinguish a definite integral from an indefinite integral a definite integral have limits of integration it has a lower limit and an upper limit an indefinite integral does not have the limits of integration so that's how you can easily distinguish between them now we know that the derivative of x cubed is three x squared according to the power rule the derivative of a variable raised to a constant is the constant times the variable raised to the constant minus one well the antiderivative of a variable raised to a constant it's going to be that variable raised to the constant plus 1 divided by the constant plus 1 and then plus c the constant of integration so that's how we could find the antiderivative of a variable raised to a constant so let's find the antiderivative of three x squared so let's rewrite the constant now the variable raised to the constant that's x squared so we're going to do is we're going to add 1 to the exponent so it's going to be 2 plus 1 and then we're going to divide by that result so it becomes 3 x cubed divided by 3 and this gives us the original function x cubed but we need to add the constant of integration plus c if you were to find the derivative of x cubed plus c x cubed will become three x squared the derivative of a constant will become zero so you end up with the same thing so always when you're looking for the indefinite integral of something or the anti-derivative you need to put the constant of integration the only time you don't need to worry about the constant of integration is if you're finding the definite integral where you get a number besides that anytime you get a function add plus c when you're looking for the indefinite integral or the antiderivative so now let's work on some problems go ahead and find the antiderivative of the following functions feel free to pause the video and work on these examples so for the first one we're just going to add one to the exponent 4 plus 1 is 5 and then we're going to divide by the result and then add the constant of integration so it's x to the fifth power over five plus c for the next one it's going to be x to the eight over eight plus c for the next one it's gonna be six and then we're gonna add one to nine that becomes ten divided by 10 plus c now this one we can reduce 6 over 10 if we divide both numbers by 2 that becomes 3 over 5. so we could say this equals 3 over 5 x to the 10 plus c for the next one it's going to be eight x to the four over four plus c and we know eight divided by 4 is 2. so this is 2 x to the 4 plus c so as you can see it's not that difficult to find the anti-derivative of a function so go ahead and try this one find the antiderivative of so let's write it this way let's say f of x is x cubed minus 4x squared plus 8x what is the antiderivative capital f of x of that function so go ahead and find the antiderivative f of x to do that we're going to find the indefinite integral of x cubed minus 4x squared plus 8x but just to show our work let's do it this way first so the antiderivative is going to be the indefinite integral of f of x dx so that's the formula and now let's replace f of x with x cubed minus 4x squared plus 8x dx so the antiderivative of x cubed is going to be x to the four over four and then for four x squared is going to be four x cubed over three and then for eight x it's eight x squared over two plus c now let's rewrite it so we're going to write it as 1 4 x to the 4 minus 4 over 3 x cubed and then 8 over 2 is 4. so we'll write that as 4x squared plus c so this is the antiderivative of the original function now let's try another problem so let's say f prime of x that is the first derivative of x i mean the first derivative of f rather let's say that's equal to eight x cubed minus six x squared plus four x minus seven so given f prime of x find the anti-derivative f of x go ahead and try that so first let's write an expression involving indefinite integral so we could say that f of x is the indefinite integral of f prime of x dx so f of x is the antiderivative of f prime of x so f of x is going to be the indefinite integral of 8x cubed minus six x squared plus four x minus seven feel free to pause the video if you wanna try this so the antiderivative of x cubed is x to the fourth over four for x squared it's x cubed over three and for x to the first power it's x squared over two and then if you have a constant like seven all you gotta do is add an x to it we'll talk more about that shortly and then finally plus c so now let's simplify our answer so the antiderivative is going to be 8 divided by 4 is 2 so we have 2 x to the fourth power 6 divided by 3 is also 2 4 divided by 2 is the same so this is the anti-derivative of our original function two x to the fourth power minus two x cubed plus two x squared minus seven x plus c so as you can see it's not too difficult to find the anti-derivative but there are other things that you need to know go ahead and find the indefinite integral of these expressions let's make this a z all right so go ahead and try those so let's start with the first one we want to find the indefinite integral of a constant 5 and we have dx in front of it so we know it's going to be 5x but let's see if we can get the answer using the power rule five is the same as five x to the zero any variable raised to the zero power is one so x to the zero is one times five which is equivalent to five now using the power rule this is going to be 5 we're going to add one to the exponent and then divide by that result so the end result is that you get 5x plus c so whenever you have a constant and you wish to integrate that constant just add a variable to it now the variable that you should add is based on what you see here either dx dy dr or dz so what is the indefinite integral of 70y so instead of 7x this is going to be 7 times y plus c so the differential dx dy it tells us what variable we should be adding to the constants the indefinite integral of 8 dr that's going to be 8 times r plus c for six z plus four it's going to be six and then z to the first power that becomes z squared over two four becomes four z and then we'll add plus c so this we could simplify it as three z squared plus four z plus c so that is the indefinite integral of six z plus four try this problem so let's say you're given the first derivative of f and let's say it's seven this should be not f of x but f prime of y so let's say it's seven y cubed minus three y plus eight and your goal is to find capital f of y how would you do it feel free to pause the video and try it so first let's write a mental outline of what we need to do here we're given the first derivative of f we need to find f so f is the antiderivative of f prime but we're not going to stop there we need to find capital f so what we need to do is we need to integrate the function two times so we have a successive integration problem so first let's integrate it the first time let's find regular f or little f so this is going to be the indefinite integral of f prime of y d y so that is the indefinite integral of 7 y cubed minus three y plus eight so the antiderivative of y to the third is going to be y to the fourth over four the antiderivative of y is y squared over two and for the constant eight there's just going to be eight y based on what we have here d y and then plus the constant c so that's f of y we don't know what the value of c is if we were given a point of f of y we could find c but we'll save that for another problem so this here is the general function of f of y so now let's find capital f of y that's going to be the indefinite integral of f of y d y so that is the indefinite integral of seven over four y to the fourth minus three over two y squared plus 8y plus c as well and we're going to have d y so the anti-derivative of this we're going to rewrite the constant and then we'll find the anti-derivative of y to the fourth so that's going to be y to the fifth over five and then minus let's rewrite the constant first so we have three over two and then the anti-derivative of y squared is going to be y to the third over three and then it's going to be plus eight times y squared over two now we have the constant of integration c so we're going to add a y to it it's going to be c times y and then plus a new constant which we'll call d so this becomes seven over twenty y to the fifth here we can cancel the threes so it's going to be minus one half y to the third plus four y squared plus c y plus d so that is the second antiderivative of f prime of y which is capital f of y by the way for those of you who want access to the full length version of this video check out the links in the description section below if you join my youtube channel membership program you can get access not only to the full-length version of this video but many other full-length versions of other videos that i have on youtube which you'll see over time so feel free to take a look at that when you get a chance now let's work on finding the indefinite integral of other types of functions what is the indefinite integral of pi d theta feel free to pause the video and work on that and what is the indefinite integral of e let's say e d s so pi is a constant let's compare to the indefinite integral of seven d y 7 is a constant and we would add the variable y to it so this will be 7 y plus c so pi is a constant and here we're going to add the variable theta to it plus c so that's the indefinite integral of pi d theta so remember pi is a constant it's 3.14 with some other numbers after that let me find out what those other numbers are so it's 3.141592654 with some other numbers so whenever you see pi it's just a constant so treat it as if it's any other number like seven or eight or negative four the same is true for e e is not a variable it's a constant e is equal to two point seven one 8 2 8 and so forth so when looking at a problem like this e is the constant s is the variable so this is going to be e times s plus c whenever you're finding the indefinite integral of something you need to be able to distinguish the constant from the variable in order to integrate it properly now let's focus on finding the indefinite integral of rational functions what is the indefinite integral of 1 over x squared to find this before you integrate it you need to rewrite the expression we need to move the variable from the bottom to the top so this is x to the negative 2. at this point we could use the power rule so we can integrate it by adding one to the exponent so it's going to be negative two plus one and then dividing by that result plus c negative two plus one that's negative one and after you integrate it you need to rewrite it again so we want to get rid of this negative exponent and we can do so by moving the variable back to the bottom so that leaves a 1 on top and then we're going to have x to the first power on the bottom which we can just leave as x the negative sign i'm going to move it to the top so the final answer is negative 1 over x plus c now let's try this one what is the indefinite integral of one over x to the fifth power dx so feel free to pause the video and try it so just like before let's move the variable to the top so we have the integral of x raised to the negative 5 dx and then let's add one to that so negative five plus one is negative four and then we're going to divide by that result next let's rewrite the expression so we're going to move the variable back to the bottom and the negative sign we're going to move it to the top so we're going to have negative 1 we have a 4 on the bottom and x and then this negative 4 becomes positive 4 plus c so negative 1 over 4x to the fourth plus c that is the indefinite integral of 1 over x to the fifth power try this find the indefinite integral of 7 over x cubed dx so like before we're going to move the variable to the top so we're going to rewrite the integral so this is 7 x to the negative 3 dx and now let's integrate so negative 3 plus 1 that's negative 2 and then we're going to divide by negative 2 and then we'll have plus c so next we're going to move the variable back to the bottom and the negative sign i'm going to move to the top so this becomes negative seven we have a two on the bottom and we have x squared as well so it's negative seven over two x squared plus c so that's how you can integrate rational functions go ahead and try this expression find the indefinite integral of 5 x to the 7 minus 9 over x squared plus 4x minus 8 dx so this is going to be 5 x to the 8 over 8. now for negative 9x squared we need to write that as negative nine x to the negative two so that's gonna be negative nine x negative two plus one is negative one divided by negative one and then for four x to the first power that will be four x squared over two and then minus eight x plus c so we can rewrite this as five over eight x to the eight plus nine the two negative signs will cancel and then we can move the x variable to the bottom so it's plus whoops it's plus 9 over x and then plus 2x squared minus 8x plus c so that is the indefinite integral or the antiderivative of this expression now what is the antiderivative of 1 over x dx what do you think we need to do here well if we were to try the same approach that we've been using for rational functions let's say if we were to rewrite the expression by moving x to the top we would get the indefinite integral of x to the negative 1 dx now using the power rule if we add 1 to negative 1 we'll get x to the zero and if we were to divide by that result we would get something that is undefined so this technique doesn't work for one over x instead this is an integral that you should commit to memory the integral of one over x dx is simply ln x so you may want to add that to your list of notes and if you recall the derivative of ln u where u is a function of x is u prime over u so the derivative of ln x is going to be u prime if u is x u prime the derivative of x is one so we get one over x so that's the anti-derivative of one over x to the first power so if the anti-derivative of one over x is l and x what is the antiderivative of one over x plus three this is simply the natural log of x plus three so based on that go ahead and find the antiderivative of the following expressions 1 over x minus 5 dx 5 over x plus 2 dx and 7 over x plus 4 dx so this is going to be the natural log of x minus 5 plus c now typically if your answer is like different than l and x normally you'll see an absolute value around x minus 5. and the reason for this is that you can have a negative number inside the natural log it's going to be like it's going to give you an error in the calculator so typically you'll see an absolute value symbol for expressions that can be negative for the next one let's move the constant to the front so think of this as 5 times the integral of 1 over x plus 2 dx so this is going to be 5 times the natural log and let's put our absolute value symbol so it's 5 ln x plus 2 plus the constant of integration for the next one let's do the same thing let's move the constant to the front so this is going to be 7 natural log x plus 4 and then plus c so that's how you can integrate rational functions where the denominator is a linear function like x plus 2 x minus 5 and so forth now here's a general formula that you can use when integrating rational functions like one over u we use a function of x it's going to be one over u prime times the natural log of u plus c now this works if and this is a big if if u is a linear function in the form ax plus b now a could be zero b could be zero well if a is zero doesn't work but b could be zero but u has to be a linear function in that form if there's like an x squared or an x cube this formula is not going to work so let me repeat that u has to be linear function if you have a square root of x or anything but x to the first power this will not work so now let's try some harder examples based on that what is the indefinite integral of one over three x plus four and let's compare that to the indefinite integral of one over x plus seven so in this case we can see that u is 3x plus 4. the derivative of u which is u prime the derivative of 3x is 3 the derivative of 4 is just nothing c u prime is 3. so this becomes 1 over 3 ln absolute value 3x plus 4. and then plus c so this fraction is a result of the derivative of that expression for the next one the derivative of x plus seven is just one here the derivative of three x plus four is three so that's why you gotta incorporate that so for the next one u is gonna be x plus seven and u prime is just gonna be one so this becomes ln absolute value x plus 7 plus c so knowing that go ahead and try these two oh by the way let's go back to what we had before let's prove that this is indeed the right answer so let's find the derivative of one third times ln three x plus four so this is going to be 1 over 3 and keep in mind the derivative of ln x or lnu rather is u prime over u so the u part is what is inside of the natural log function and so that's three x plus four u prime the derivative of three x plus four is just three and the derivative of c is zero and so when we cancel the threes we're going to start with our original function which is what we had here so that shows you that the process works you