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Main content
Current time:0:00Total duration:3:35
FUN‑6 (EU)
FUN‑6.D (LO)
FUN‑6.D.1 (EK)

Video transcript

what we're going to do in this video is give ourselves some practice in the first step of you substitution which is often the most difficult for those who are first learning it and that's recognizing when you substitution is appropriate and then defining and appropriate you so let's just start with an example here so let's say we want to take the indefinite integral of 2x plus 1 times the square root of x squared plus X DX does u substitution apply here and if it does how would you define that you pause the video and try to think about that well we just have to remind ourselves that the U that u substitution is really trying to undo the chain rule if we remind ourselves what the chain rule tells us it says look if we have a composite function let's say F of G of X F of G of X and we take the derivative of that with respect to X that that is going to be equal to the derivative of the outside function with respect to the inside function so f prime of G of X times the derivative of the inside function times the derivative of the inside function and so u substitution is all about well do we see a pattern like that inside the integral do we see a potential inside function a G of X where I see its derivative being multiplied well we see that over here if I look at x squared plus X if I make that the U what's the derivative of that well the derivative of x squared plus X is 2x plus 1 so we should make that substitution if we say u is equal to x squared plus X then we could say D u DX the derivative of U with respect to X is equal to 2x plus 1 if we treat our differentials like variables or numbers we can multiply both sides by DX which is a little bit of hand wavy mathematics but it's appropriate here so we could say 2x plus 1 times DX and now what's really interesting is here is we have our u right over there and notice we have our 2x plus 1 DX in fact it's not conventional to see an integral rewritten the way I'm about to write it but I will I could rewrite this integral you should really view this is the product of three things oftentimes people just view the DX is somehow part of the integral operator but you could rearrange it this would actually be legitimate you could say the integral of the square root of x squared plus x times 2x plus 1 DX and if you wanted to be really clear you could even put all of those things in parenthesis or something like that and so here this is our U and this right over here is our D U and so we could rewrite this as being equal to the integral of the square root of U because x squared plus X is U times D U which is much easier to evaluate if you are still confused there you might recognize it if I rewrite this as u to the one-half power because now we could just use the reverse power rule to evaluate this and then we would have to undo the substitution once we figure out what this antiderivative is we then would then reverse substitute the X expression back in for the U
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