Need help with the substitution method of integrating... Stackers (I tried to be nifty), I'm currently in a Calculus II course, and we're working through the very basic steps of integrating. I'm really struggling with multiple portions of the substitution method that incorporates the use of _u_ and _du_. I'm looking at this problem: $$\int (x^6-3x^2)^4(x^5-x)dx$$ Now comes my dilemma. I just cannot, for the life of me, understand how the substitution method would work with this equation. I just need someone to break it down for me a bit better than the book is, I'm supposed to find a value of _u_ that helps the equation look like: $$\int u^4du$$ or something, right? If someone could just step through this problem with me, it'll help me greatly with the other 50 questions I have to do with the same concept. Is there any sort of fool-proof analysis that I can do to each problem to know where and how to assign the values of _u_ and _du_? Thanks for your help.

You don't like the power of something complicated (in particular, you do _not_ want to have to expand the power). So you want the inside of the power to be $u$. (This is a good thing to try in other problems; depending on the setup, it may or may not work.) Then $u'=6x^5-6x$. You pretty much have this already, except for the $6$, but you can bring in constant multiples as you wish, provided you balance them. Specifically, you can write $\int (x^6-3x^2) (x^5-x) dx = \frac{1}{6} \int (x^6-3x^2) (6x^5-6x) dx$. Then with $u=(x^6-3x^2)$, your integral is $\frac{1}{6} \int u^4 du$.