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we're faced with the indefinite integral of X plus 3 times X minus 1 to the fifth DX now we could solve this by literally multiplying out what X minus 1 of the 5th is maybe using the binomial theorem that would take a while and then we'd multiply that times X plus 3 and we'd end up with some polynomial and we could take the antiderivative that way or we could maybe make a substitution here that could simplify this expression here make it something that's a little bit easier to take the antiderivative of and this isn't going to be kind of the more traditional use substitution or just set u equal to something and see if it's derivative is there but it's a kind of a form of u substitution where we do set u equal to something and see if it simplifies our expression in a way that I guess simplifies it so let's try things out so we had this X minus 1 to the fifth that would be a pain to expand out it would be nice if this was just a U to the fifth so let's just set this let's just set this to be equal to u so let's set U is equal to X minus 1 and in that case D U is equal to DX we could write D U DX is equal to 1 derivative of X derivative negative 1 is just 0 and these two these two are completely equivalent statements and so if we did that how could we rewrite this entire expression well it would be equal to the integral of well we have X plus 3 right over here this is neither just you nor is it D u so let's think about how what we could do here well we could say if u is equal to X minus 1 if u is equal to X minus 1 we could add 1 to both sides of this equation and we could say u plus 1 is equal to X and so for X we can substitute that with u plus 1 so let's do that so we're kind of back substituting in for X so X is equal to u plus 1 X is equal to u plus 1 and I'm just trying to see if I can do something to simplify this expression so X is U plus 1 X is U plus 1 then we have our plus 3 there plus 3 times X minus 1 to the fifth X minus one was you that's the simplification we wanted to make so times U to the fifth power and DX is the same thing as D u so D u now did this get us anywhere can we simplify this to a form that it's easy to take the antiderivative of well it looks like it did let's see we can rewrite this as this expression right over here is just u plus four so it's u plus four times U to the fifth times U to the fifth I'll do it all in one color now D U and the reason why the simplified thing is the the way to simplify things is taking X minus one of the fifth that would be a really hard thing to expand but do to the fifth it's just you to the fifth and then it just changed this X plus 3 into a u plus four which is still a pretty straightforward expression and now we can just distribute the U to the fifth so we are left with U to the sixth power plus four plus four u to v D U and this is a pretty straightforward thing to take the antiderivative of now you might be saying hey Sal what you know this is how do you know to set u equal to B that and often times those integrations going to take a little bit of trial and error there's a certain bit of an art to it but here the realization was well is X minus one of the fifth it could be really complicated maybe you know the fifth might make a little bit simpler and that did it just happened to work you could have tried U is equal to X plus three but it wouldn't have simplified it as nicely as you to X minus one did but let's finish with this end this integral right over here so this is going to be equal to the antiderivative of U to the sixth well that's just you to the 7 over 7 plus the antiderivative of U to the fifth that's U to the sixth over six but we have the four out here so it's four times U to the six over six and then we have a plus C and this we the four six is the same thing as 2/3 so we can rewrite this whole thing is equal to U to the seventh over seven plus 2/3 you to the six plus C and now we just have to undo our u substitution U is equal to X my one so this is going to be equal to X minus 1 to the 7 or 7 plus 2/3 times X minus 1 to the sixth plus C and we're all done we were able to take a fairly hairy or what could have been a hairy thing if we had to expand this out and we were able to take the antiderivative by doing a little bit of this u substitution and you back substitution