Question

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

Answer 1

$Let\ x^3 = t$

$∴ 3x^2 \ dx = dt$

$⇒ ∫\frac {x^2}{1-x^6} \ dx = \frac 13 ∫\frac {dt}{1-t^2}$

$=\frac 13[\frac 12\ log\ |\frac {1+t}{1-t}|]+C$

$=\frac 16\ log\ |\frac {1+x^3}{1-x^3}|+C$