Integrate the function: $\frac{x^{2}}{1-x^{6}}$

Let $x^{3} = t$.
Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$,which implies $x^{2} dx = \frac{1}{3} dt$.
Substituting these into the integral:
$\int \frac{x^{2}}{1-x^{6}} dx = \int \frac{1}{1-(x^{3})^{2}} (x^{2} dx) = \frac{1}{3} \int \frac{dt}{1-t^{2}}$.
Using the standard integral formula $\int \frac{1}{a^{2}-x^{2}} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,where $a=1$:
$= \frac{1}{3} \left[ \frac{1}{2(1)} \log \left| \frac{1+t}{1-t} \right| \right] + C$
$= \frac{1}{6} \log \left| \frac{1+x^{3}}{1-x^{3}} \right| + C$,where $C$ is an arbitrary constant.