in this video we're going to go over a few indefinite integral problems so what is the integral of 4 dx what is the answer for this problem the anti-derivative of a constant all you need to do is just add an x to it this is going to be 4x and you also need to add a a c value anytime you integrate a function there's always going to be a constant that you need to add to it now the derivative of 4x is 4. the derivative of any constant is 0. so that's why you always need to add the constant so what about let's say the antiderivative of pi let's say it's d y instead of dx all we need to do is add a y variable to it it's going to be pi times y plus c now what about the antiderivative of e dz e is a constant so it's just going to be e times z plus c now the next type of problem that you're going to see is when you need to integrate a variable raised to a constant let's say x raised to the n this is equal to x raised to the n plus one divided by n plus one plus c so for example let's say if we wish to find the antiderivative of x squared dx this is equal to x to the third divided by three plus c the antiderivative of x to the third is x to the fourth divided by four plus c try this one what is the antiderivative of eight x cubed so focus on x to the third power the eight is just gonna come along for the ride the anti-derivative of x to the third is x to the fourth divided by four and now we can simplify eight divided by four is two so the final answer is 2x to the fourth plus c try this one what is the antiderivative of 5 x to the sixth power so using the same technique let's add one to the exponent and then divide by that result so this is the answer now what about this expression what's the anti-derivative of 7x dx well there is an invisible one so we can use the same technique if we add one it's going to be 2 and then divide by it plus c so the anti-derivative of 3x is simply x squared divided by two plus c what if we have a polynomial function x squared minus five x plus six so you need to integrate each one separately the anti-derivative of x squared is x to the third divided by three and for 5x is going to be 5x squared divided by 2 and then if you have a constant just add a variable to it so this is the answer for each of these problems before i begin feel free to pause the video and work on it so try this one 4x cubed plus 8x squared minus 9 dx so the anti-derivative of x cubed is x to the fourth divided by 4 and for x squared is x to the third divided by 3 and for the constant simply add an x to it and if you can reduce it go ahead and do so four divided by four is one so it's x to the fourth plus eight thirds x cubed minus nine x plus c so this is the answer now what about square root functions what is the anti-derivative of the square root of x if you see a question like this rewrite it this is the same as x to the one-half now we need to add one to the exponent one-half plus one is the same as 1 half plus 2 over 2 which is 3 over 2. so this is going to be x to the 3 halves and instead of dividing it by 3 halves you can multiply by the reciprocal which is two-thirds so you can write the final answer as two-thirds square root x-cubed plus c keep in mind there is an invisible 2 here here's another one that you can try what is the anti-derivative of the cube root of x to the fourth so go ahead and pause the video and work on this example now the first thing that we need to do is rewrite it you can rewrite it as x to the four thirds now four thirds plus one is the same as four over three plus three over three which is seven over three so once we add one to the exponent it's just going to be x to the seven thirds instead of dividing it by seven over three let's multiply it by three over seven and then add the plus c constant so the final answer is three over seven cube root x to the seventh power plus c now what about this one what's the antiderivative of 3x minus 1 squared dx what would you do in this problem best thing to do is to foil this expression three x minus one squared is three x minus one times three x minus one three x times three x is nine x squared and then three x times negative one that's negative three x negative one times three x is also negative three x and finally we have negative one times negative one which is plus one so now let's combine like terms negative 3x plus negative 3x is negative 6x so this is what we now have the antiderivative of x squared is x cubed divided by three and for x to the first power is x squared divided by two and for the constant add an x to it so now let's simplify nine divided by three is three six divided by two is three and so this is the answer let's try this one two x plus one times x minus two dx so just like the last example we need to foil first 2x times x is 2x squared and then 2x times negative 2 that's negative 4x 1 times x is x and then 1 times negative 2. now negative four x plus x is negative three x so now let's integrate it this is going to be two x cubed divided by three minus three x squared divided by two minus two x plus c and this is the answer we can't really simplify it so we're gonna leave it like that let's try this one x to the fourth plus six x cubed divided by x dx what should we do here what would you do in this problem now if you have a fraction with a single term in the bottom separate it into two smaller fractions so x to the fourth divided by x is x cubed six x cubed divided by x is six x squared and now as you can see it's fairly easy to find the anti-derivative so this is going to be x to the fourth divided by four plus six x cubed divided by three plus c which we can write as one fourth x to the fourth plus 2 x cubed plus c what about this one what's the anti-derivative of 1 over x squared dx how can we integrate this function for fractions like this you want to rewrite it if you move the x variable from the bottom to the top the exponent is going to change sign it's going to change from positive 2 to negative 2. and now you can use the power rule so if we add 1 to negative 2 and then divide by that result this is going to be x to the negative 1 divided by negative 1 plus c now we can rewrite it we can move the x back to the bottom so the final answer is negative one over x plus c now let's try another one like that try this one one over x cubed so first let's rewrite it this is x to the negative three dx and now let's add one to the exponent negative three plus one is negative two divide by that result and you should get this now let's rewrite it so we have a negative in front we have a two on the bottom let's keep it there and we're going to move the x to the bottom as well so the negative 2 is going to change to positive 2. so it's negative 1 over 2 x squared plus c now let's try this one 5 divided by x to the fourth so first let's rewrite it this is 5 x to the negative 4 and then let's add 1 to the exponent so negative 4 plus 1 is negative 3 and then divide by negative 3 and then we'll move the x back to the bottom so it's negative 5 divided by 3 x cubed plus c now what is the anti-derivative of one over x if we try to rewrite it and if we add one to the exponent this is going to be negative one plus one is zero and if you have zero on the bottom it's undefined so this is not going to work for now you just want to know that the anti-derivative of 1 over x is ln x likewise let's say if you want to find the anti-derivative of one over x minus three this is simply ln x minus three the anti-derivative of one over let's say uh x plus four this is just going to be ln x plus 4. now what about this one what's the anti-derivative of 5 over x minus 2 if you want you can move the constant to the front so this expression is equivalent to five times one over x minus two dx and this portion is equal to ln x minus two and just multiply by five so this is the answer now let's move on to exponential functions what is the anti-derivative of e to the four x if it's e to the some number x to the first power here's what you need to do it's just going to be e to the 4x divided by the derivative of 4x plus c if it's like x squared on top or x cube or something else it won't work it only works if the exponent is a linear function so if you want to find the derivative of e to the 5x it's simply going to be e to the 5x divided by 5. so what about e to the x it's going to be e to the x divided by the derivative of x which is 1 plus c so the anti-derivative of e to the x stays the same it's just e 3 x now let's try a few more examples try these two eight e to the two x and also twelve e to the three x so this is to be 8 e to the 2x divided by the derivative of 2x which is 2 and that reduces to 4 e to the 2x plus c for the last one this is going to be 12 e to the 3x divided by 3 plus c and that reduces to 4 e to the 3x plus c now let's move on to trig functions what's the antiderivative of cosine x dx now think backwards the derivative of what function is cosine the derivative of sine is cosine so the antiderivative of cosine is positive sine now what is the anti-derivative of sine the derivative of cosine is negative sine so the derivative of negative cosine is positive sine which means the anti-derivative of positive sign is negative cosine so let's say if we want to find the anti-derivative of cosine 3x this is going to be sine 3x the angle has to stay the same but then divided by the derivative of 3x which is 3. this works is only if you have a linear function on the inside can you use this technique so for example the antiderivative of cosine 7x is simply sine 7x divided by 7 plus c now what is the antiderivative of 14 sine two x so this is just going to be fourteen the anti-derivative of sine is negative cosine two x but divided by two so we can reduce that to negative seven cosine two x plus c fourteen divided by two is seven let's try this one 6 sine 3x dx the antiderivative of sine is negative cosine 3x but divided by 3. so the final answer is negative 2 cosine 3x plus c 6 divided by 3 is 2. now what is the antiderivative of secant squared dx the derivative of tangent and secant squared so the anti-derivative of secant squared is simply tangent x so if we want to find the anti-derivative of 8 secant squared 4x this is going to be 8 times tangent 4x but divided by 4 plus c which becomes 2 tangent 4x plus c now what is the antiderivative of secant x tangent x the derivative of secant is secant tangent so this is equal to just secant x plus c so if we have the antiderivative of 12 secant 3x tangent 3x this is equal to 12 secant 3x divided by 3 plus c which is just uh 4 secant 3x plus c now what if you were to see an expression that looks like this the what is the antiderivative of x squared sine x cubed dx what would you do in a problem like this now there's a technique called u substitution and you want to replace all the x variables with u variables i'm going to make u equal to x cubed the reason being is the derivative of x cube is 3x squared and the 3x squared in this expression can cancel with the x squared in that expression which is what we want now in the next step solve for dx du divided by 3x squared is equal to dx so what we're going to do is replace x cubed with u and dx with du divided by 3x squared so this is going to be x squared sine u d u over 3x squared so notice that the x squared cancels and let's take this constant and move it to the front so we now have is one third anti-derivative sine udu and we know what the anti-derivative of sine is it's negative cosine so this is negative one-third cosine u plus c now at this point all you need to do is replace the u variable with what it was in the beginning x cubed so the final answer is negative one third cosine x cubed plus c let's try another u substitution problem try this one do you think we should make u equal to x squared plus 3 or 5x notice that the derivative of x squared will give you 2x which can cancel the x and 5x so you want to make u equal to x squared plus 3x so du is going to be 2x dx and then solve for dx dx is du divided by 2x now we need to replace x squared plus 3 with u and dx with du over 2x so this is going to be 5x raised to the u or times u raised to the fourth power and then times d u divided by 2x so we can cancel the x variable and the constant 5 over 2 let's move it to the front so let's put it on the left side of the integral so this is 5 over 2 u to the fourth du now we can use the power rule so if we add one to the exponent it's going to be u to the fifth divided by five plus c so notice we can cancel these fives so what we now have is one half u to the fifth plus c which is let's make some space at this point we can replace u with x squared plus three so the final answer is one half x squared plus three raised to the 5th power plus c so this is it now let's try this problem what is the anti-derivative of tangent x you can either know this answer or you could find a way to get the answer so what can we do now tangent is sine divided by cosine so in this form we could use u substitution let's replace u with cosine x if we do that the derivative of cosine is going to be negative sine and if we solve for dx it's going to be d u divided by negative sine notice that the sign variable will cancel so let's replace cosine with the u variable and let's replace dx with du divided by negative sign so the expression that we now have is the anti-derivative of negative one over u if you recall the anti-derivative of one over x is l and x so for one over u it's ln of u now we can replace the u variable with cosine so we now have is negative ln cosine x now there's a one in front of here a property of natural logs allows you to take the coefficient and move it inside of ln so this is going to be ln cosine x to the negative 1 power plus c which is the same as ln 1 divided by cosine x now 1 divided by cosine is secant so the final answer is ln secant x plus c that is the antiderivative of tangent x now what if you were to see this what is the anti-derivative of x cosine x dx we can't really use u substitution here the derivative of cosine is negative sine that's not going to cancel with the x so u substitution won't work there's something else called integration by parts and here's the formula the integration of u dv is equal to uv minus the anti-derivative of v d u let's make u equal to x so this is the u part dv we're going to make it equal to cosine x dx now we need to find d u and v d u is the derivative of u so the derivative of x is one v is the anti-derivative of d v the anti-derivative of cosine is sine so this is going to be u times v that's uh x times sine x minus the antiderivative of v d u which is simply sine x oh let's not forget d x is here d u is 1 dx and udv is basically the original function now all we need to do is find the antiderivative of sine the antiderivative of sine is negative cosine so we have the final answer it's going to be x sine x plus cosine x plus c let's try another integration by parts problem try this one x e to the 4x dx so you want to make u equal to x because when you find d u the x is going to disappear d is going to be 1 dx now we're going to make dv equal to e to the 4x because we know how to find the anti-derivative of e to the 4x e to the x and sine x they are basically repeating functions the anti-derivative of e to the 4x is e to the 4x divided by 4. so using the formula anti-derivative udv is uv minus anti-derivative to u so u v is basically uh x times one fourth e to the four x so that's one fourth x e to the four x minus the anti-derivative of v d u which is simply 1 4 e to the 4 x dx so the anti-derivative of e to the 4 x is just e to the four x divided by four and then plus c so now we can write the final answer which is one fourth x e to the four x minus 4 times 4 is 16 so 1 over 16 e to the 4x plus c so that's integration by parts let's try this problem what is the antiderivative of 4 divided by 1 plus x squared dx now in this problem we can't really use u substitution and we can't use integration by parts so what can we do here now another technique that you can use is trigonometric substitution it helps to know that 1 plus tangent squared is equal to secant squared so notice the expression 1 plus x squared that is an indication that we should make x equal to tangent theta if x is tangent theta x squared is going to be tangent squared and dx is the derivative of tangent which is secant squared d theta so let's replace x squared with tangent squared and let's replace dx with secant squared theta d theta so now 1 plus tan squared we know is secant squared and the secant squareds cancel in this problem so what we now have is the antiderivative of 4d theta the antiderivative of 4dx is just 4x so 40 theta is simply 4 theta plus c so now we need to replace data with something in terms of x if x is equal to tangent theta then the inverse tangent of x is equal to theta whenever you're dealing with an inverse function you need to switch x and y in this case x and theta so the final answer is 4 inverse tangent of x plus c we just need to replace theta with inverse tan so this is it let's try this one now sine squared plus cosine squared is equal to one so one minus sine squared is equal to cosine squared so this is another trigonometric substitution and based on that identity we want to make x equal to sine so if x is equal to sine theta x squared is sine squared theta and dx is the derivative of sine is going to be cosine theta d theta so let's replace x squared with sine squared and let's replace dx with cosine theta d theta so we know that one minus sine squared is equal to cosine squared and the square root of cosine squared is simply cosine theta so at this point we can cancel the cosine function so the expression that we have now is simply the anti-derivative of three d theta which is equal to three theta plus c so now our last step is to replace theta with something so if x is equal to sine theta what we need to do is take the inverse sine of both sides the inverse sine of sine theta is simply theta these two cancel so inverse sine x is equal to theta so the final answer is 3 inverse sine x plus c