<s>[INST] 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.
 [/INST] Set $t=x^3+1$ and $dt=3x^2dx$ 
Therefore $\color{red}{ x^2dx=\frac{du}{3}}$ 
$$=\frac 1 3\int \frac 1 t dt =\frac{\ln t}{3}+\mathcal C=\boxed{\color{blue}{\frac 1 3 \ln (x^3+1)+\mathcal C}} $$
hope it helps
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