Subsitution - integrals 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.

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