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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 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.2011-11-11

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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)

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

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