# 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?

• well, have you an idea for $u$? hint: you are looking for some function $u$ such that the integral looks like (not exactly) $\int (u(x))^4 u'(x) \; \mathrm d x$. – user251257 Sep 8 '15 at 23:16
• No, that's what I'm struggling with here. I don't know how to determine what portion of the original equation to designate as u. If that makes sense, I need to figure that part out. – MLin16 Sep 8 '15 at 23:19
• your integral looks like $\int (...)^4(...) \; \mathrm d x$ ... So what would be a candidate for $u$? – user251257 Sep 8 '15 at 23:20
• Look for something in the original equation where if you make the substitution $u = f(x)$, you'll also be able to find the derivative $u'=f'(x)$ in the original equation as well. – NoseKnowsAll Sep 8 '15 at 23:20
• I don't know how to write on stackexchange, so I don't know how to use the integral symbol in replies without it going centered and looking weird. Sorry. But, in a sense, my equation should look like: $$\int(f(x)^4(f'(x))dx$$ or, am I wrong again? – MLin16 Sep 8 '15 at 23:22

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 $\frac{1}{6} \int u^4 du$.