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.

A more rigoures approach.
$$ \int \frac{x^2}{x^3+1} dx $$
$$ u = x^3 +1 $$
$$ \frac{du}{dx}  = 3x^2 $$
so:
$$ \frac{1}{3} \int \frac{3x^2}{x^3+1} dx =  \frac{1}{3}  Ln |x^3 +1| + K$$
