Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.

Let $u=x^3+1$ then you have your $du=3x^2dx$, then from your original equation $x^2dx$ is just equal to $\frac{du}{3}$. Now solving for the whole equation that just equal to $\int\frac{du}{3u}$.
$\int\frac{du}{3u}=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}lnu+ C= \frac{1}{3}ln(x^3+1)+C.$
