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 four the derivative of any constant is zero so that's why you always need to add the constant so what about let's say the anti-derivative of pi let's say it's dy instead of dx all you need to do is add a y variable to it it's going to be pi * y + c now what about the anti-derivative of e dz e is a constant so it's just going to be E * 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 + 1 / n + 1 + c so for example let's say if we wish to find the anti-derivative of x^2 dx this is equal to x 3r / 3 + c the anti-derivative of x 3r is x 4 / 4 + c try this one what is the anti-derivative of 8x cub so focus on x^ the 3 power the 8 is just going to come along for the ride the anti-derivative of x 3r is x 4 / 4 and now we can simplify 8 / 4 is 2 so the final answer is 2x 4 + c try this one what is the anti-derivative of 5x^ 6 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 two and then divide by it plus c so the anti-derivative of 3x is simply 3x^2 / 2 + c what if we have a polomial function x^2 - 5x + 6 so you need to integrate each one separately the anti-derivative of x^2 is x 3r / 3 and for 5x is going to be 5x^2 / 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 + 8 x^2 - 9 dx so the anti-derivative of x is x 4 / 4 and for x^2 it's x 3r / 3 and for the constant simply add an x to it and if you can reduce it go ahead and do so 4 / 4 is 1 so it's x 4 + 8/3 xqub - 9 x + 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 12 now we need to add one to the exponent 12 + 1 is the same as 12 + 2 over 2 which is 32 so this is going to be x to the 3s and instead of dividing it by 3 halves you can multiply by the reciprocal which is 2/3 so you can write the final answer as 2/3 x cub + c keep in mind there is an invisible two here here's another one that you can try what is the anti-derivative of the cube root of x 4th 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^ 4/3 now 4/3 + 1 is the same as 4 over 3 + 3 over 3 which is 7 over 3 so once we add 1 to the exponent it's just going to be x^ 7/3 instead of dividing it by 7 over 3 let's multiply it by 3 over 7 and then add the plus c constant so the final answer is 37 cube root x^ the 7th power plus c now what about this one what's the anti-derivative of 3x - 1^ 2 dx what would you do in this problem the best thing to do is to foil this expression 3x - 1^ 2 is 3x - 1 * 3x - 1 3x * 3x is 9 x^2 and then 3x * -1 that's -3x -1 * 3x is also -3x and finally we have -1 * 1 which is + 1 so now let's combine like terms -3x + -3x is -6x so this is what we now have the anti-derivative of x^2 is x cub / 3 and for x to the 1st power is x^2 / 2 and for the constant add an x to it so now let's simplify 9 / 3 is 3 6 / 2 is 3 and so this is the answer let's try this one 2x + 1 * x - 2 dx so just like the last example we need to foil first 2x * x is 2x^2 and then 2x * -2 that's -4x 1 * x is x and then 1 * -2 now -4x + x is -3x so now let's integrate it this is going to be 2 x cub / 3 - 3x^2 / 2 - 2x + c and this is the answer we can't really simplify it so we're going to leave it like that let's try this one x 4 + 6 x cub / 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 4 / x is x cub 6 x cub / x is 6 x^2 and now as you can see it's fairly easy to find the anti-derivative so this is going to be x 4 / 4 + 6 xub / 3 + c which we can write it as 1/4 x 4 + 2 x cub + c what about this one what's the anti-derivative of 1x^2 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 pos2 to -2 and now you can use the power rule so if we add one to -2 and then divide by that result this is going to be x^1 /1 + c now we can rewrite it we can move the x back to the bottom so the final answer is -1x + c now let's try another one like that try this one 1x cub so first let's rewrite it this is x to the -3 dx and now let's add one to the exponent -3 + 1 is -2 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 in the bottom let's keep it there and we're going to move the x to the bottom as well so the -2 is going to change to pos2 so it's -1 / 2x^2 + c now let's try this one 5 / x 4th so first let's rewrite it this is 5x^ -4 and then let's add 1 to the exponent so4 + 1 is -3 and then divide by3 and then we'll move the x back to the bottom so it's -5 / 3 x cub + c now what is the anti-derivative of 1 /x if we try to rewrite it and if we add one to the exponent this is going to be 1 + 1 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 /x is ln x likewise let's say if you want to find the anti-derivative of 1 /x - 3 this is simply ln xus 3 the anti-derivative of 1 / let's say uh x + 4 this is just going to be ln x + 4 now what about this one what's the anti-derivative of 5 x - 2 if you want you can move the constant to the front so this expression is equivalent to 5 * 1x - 2 dx and this portion is equal to ln x - 2 and just multiply by 5 so this is the answer now let's move on to exponential functions what is the anti-derivative of e to 4x if it's e to the some number x to the 1st power here's what we need to do it's just going to be e 4x / the derivative of 4x plus c if it's like x^2 on top or x cub 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^ 5x it's simply going to be e 5x / 5 so what about e to x it's going to be e x / 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 to the x now let's try a few more examples let's try these two 8 e to the 2x and also 12 e to the 3x so this is going to be 8 e 2x / the derivative of 2x which is 2 and that reduces to 4 e 2x + c for the last one this is going to be 12 e^ 3x / 3 + c and that reduces to 4 e^ 3x + c now let's move on to trig functions what's the anti-derivative of cosine x dx now think backwards the derivative of what function is cosine the derivative of s is cosine so the anti-derivative of cosine is positive sign now what is the anti-derivative of s the derivative of cosine is negative sign so the derivative of negative cosine is positive sign which means the anti-derivative of positive s is negative cosine so let's say if we want to find the anti-derivative of cosine 3x this is going to be sin 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 anti-derivative of cossine 7x is simply sin 7x / 7 + c now what is the anti-derivative of 14 sin 2x so this is just going to be 14 the anti-derivative of s is negative cosine 2x but divided by 2 so we can reduce that to -7 cossine 2x + c 14 / 2 is 7 let's try this one 6 sin 3x dx the anti-derivative of s is cosine 3x but / 3 so the final answer is -2 cosine 3x + c 6 / 3 is 2 now what is the anti-derivative of secant^ 2 dx the derivative of tangent is squ so the anti-derivative of squ is simply tangent x so if we want to find the anti-derivative of 8^ 4x this is going to be 8 * tangent 4x but / 4 + c which becomes 2 tangent 4x + c now what is the anti-derivative of secant x tangent x the derivative of secant is secant tangent so this is equal to just sec x plus c so if we have the anti-derivative of 12 secant 3x tangent 3x this is equal to 12 secant 3x / 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 anti-derivative of x^2 sin x cub 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 cub the reason being is the derivative of x cub is 3x^2 and the 3x^2 in this expression can cancel with the x^2 in that expression which is what we want now in the next step solve for dx du / 3x^2 is equal to dx so what we're going to do is replace x cub with u and dx with du / 3x^2 so this is going to be x^2 sin u du over 3x^2 so notice that the x^2 cancels and let's take this constant and move it to the front so what we now have is 1/3 anti-derivative sin u du and we know what the anti-derivative of s is it's negative cosine so this is 1/3 cosine u + c now at this point all you need to do is replace the u variable with what it was in the beginning x cub so the final answer is 1/3 cossine x cub + c let's try another u substitution problem try this one do you think we should make u= x^2 + 3 or 5x notice that the derivative of x^2 will give you 2x which can cancel the x and 5x so you want to make u= x^2 + 3x so du is going to be 2x dx and then solve for dx dx is du / 2x now we need to replace x^2 + 3 with u and dx with du over 2x so this is going to be 5x raised to the u or time u raised to the 4th power and then time du / 2x so we can cancel the x variable and the constant 5 / 2 let's move it to the front so let's put it on the left side of the integral so this is 52 u 4 du now we can use the power rule so if we add one to the exponent it's going to be u 5 / 5 + c so notice we can cancel these fives so what we now have is 12 u to 5th + c which is let's make some space at this point we can replace u with x^2 + 3 so the final answer is 12 x^2 + 3 raised to the 5th power + 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 sin / 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 sign and if we solve for dx it's going to be du / s notice that the sign variable will cancel so let's replace cosine with the u variable and let's replace dx with du / s so the expression that we now have is the anti-derivative of -1 / u if you recall the anti-derivative of 1 /x is ln x so for 1 over u it's ln of u now we can replace the u variable with cosine so what we now have is ln cossine 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 cossine x to the -1 power plus c which is the same as ln 1 / cosine x now 1 / cosine is secant so the final answer is ln see x + c that is the anti-derivative 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 sign 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 du 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 du and v du is the derivative of u so the derivative of x is 1 v is the anti-derivative of dv the anti-derivative of cosine is s so this is going to be u * v that's uh x * sin x minus the anti-derivative of v du which is simply sinx oh let's not forget dx is here du is 1 dx and udv is basically the original function now all we need to do is find the anti-derivative of s the anti-derivative of s is negative cosine so we have the final answer it's going to be x sin x plus cosine x plus c let's try another integration by parts problem try this one x e 4x dx so you want to make u equal to x because when you find du the x is going to disappear d is going to be 1 DX now we're going to make DV equal to E 4X because we know how to find the anti-derivative of E 4X e to X and sin X they're basically repeating functions the anti-derivative of E 4X is E 4X / 4 so using the formula anti-derivative UV is UV minus anti-derivative V du so uv is basically uh x * 1/4 e 4x so that's 1/4 x e 4x minus the anti-derivative of v du which is simply 1/4 e 4x dx so the anti-derivative of e 4x is just e 4x / 4 and then plus c so now we can write the final answer which is 1/4 x e 4xus 4 * 4 is 16 so 1 16 e 4x + c so that's integration by parts let's try this problem what is the anti-derivative of 4 / 1 + x^2 dx now in this problem we can't really use you 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 + tangent squ is equal to squ so notice the expression 1 + x^2 that is an indication that we should make x= tangent theta if x is tangent theta x^2 is going to be tangent and dx is the derivative of tangent which is dt so let's replace x^2 with tang^ 2 and let's replace dx with squ theta d theta so now 1 + tan^ 2 we know is 2 and the squares cancel in this problem so what we now have is the anti-derivative of 4 d theta the anti-derivative of 4 dx is just 4x so 40 theta is simply 4 + c so now we need to replace theta 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 in tangent of x plus c we just need to replace theta with inverse tan so this is it let's try this one now sin^ 2 + cosine^ 2 is equal to 1 so 1 - sin^ 2 is equal to cosine^ square so this is another trigonometric substitution and based on that identity we want to make x= x= sin theta x^2 is sin^ 2 and dx is the derivative of s it's going to be cosine theta d theta so let's replace x^2 with sin^ 2 and let's replace dx with cosine theta d theta so we know that 1 - sin^ 2 is equal to cosine^ 2 and the square root of cosine^ 2 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 3 d theta which is equal to 3 theta plus c so now our last step is to replace theta with something so if x is equal to sin theta what we need to do is take the inverse sign of both sides the inverse sign of sin theta is simply theta these two cancel so inverse sin x is equal to theta so the final answer is 3 inverse sin x + c