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I am attempting to solve this integral using substituion $\int (x^2 +1) (x^3 +3x)^4dx$ I make $u=x^3+3x$ and then made $dx=du/(3x^2 + 3)$ I then got $1/3 \int (x^3+3x)^4$ I have no idea what to do now.

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You are working the differentials backwards. You want to notice that $du = (3x^2+3)dx = 3(x^2+1)dx$, so $\frac{1}{3}du = (x^2+1)dx$. Oh, look! There's an $(x^2+1)\,dx$ in the original integral... – Arturo Magidin Nov 11 '11 at 1:40
I don't see what I did backwards, the du/dx part? – Jordan Nov 11 '11 at 1:42
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This isn't really an answer so I'll make it a comment. When doing the problems, keep in mind that the principle behind the substitution is essentially the same behind that of the chain rule in derivatives, just the reverse. So if you understand one, use that to understand the other. – analysisj Nov 11 '11 at 1:44
You did it backwards because you have $dx$ expressed in terms of both $du$ and $x$; you "solved for $dx$" instead of solving for $du$, which what you normally want to do. – Arturo Magidin Nov 11 '11 at 1:45

1 Answer

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Your substitution $u=x^3+3x$ is a good idea as it yields $\frac{du}{dx}=3x^2 + 3$

Substituting in you get:

$\int (x^2 +1) (x^3 +3x)^4dx$

$= \frac{1}{3} \int (3x^2 + 3)(x^3 + 3x)^4 dx$

$= \frac{1}{3} \int u^4 du$ (this is where you seemed to have gone wrong by converting to du yet leaving the integral in terms of x)

$= \frac{1}{15} u^5 + c$

$= \frac{1}{15} (x^3 + 3x)^5 + c$

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