Question

Evaluate:
$\int \frac{x^2}{1-x^6}dx$

Answer 1

$\int \frac{x^2}{1-x^6}dx$

$=\frac{1}{3}\int \frac{3x^2}{1-{\left(x^3\right)}^2}dx$

Let $x^3=t3x^{2}dx=dt$

$\therefore \frac{1}{3}\int \frac{dt}{1-t}=\frac{1}{3}(-\ln |1-t|)$

$\because t=x^3$

Therefore,

$\int \frac{x^2}{1-x^6}=\frac{-1}{3}\ln |1-x^3|$