Question

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

Answer 1

Let $x^3=t$
$\therefore 3x^{2}dx=dt$

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

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

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