Prove

Let $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$\Rightarrow \int \frac{x^{2}}{1-x^{6}} d x=\frac{1}{3} \int \frac{d t}{1-t^{2}}$

$=\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]+\mathrm{C}$

$=\frac{1}{6} \log \left|\frac{1+x^{3}}{1-x^{3}}\right|+\mathrm{C}$