Question

Evaluate the following:
$\int \frac{x^2{\tan }^{-1}{x}^3}{1+x^6}dx$

Answer 1

Let ${\tan }^{-1}{x}^3=t$

$\Rightarrow \frac{1}{1+x^6}\times 3x^{2}dx=dt$

$\Rightarrow \int x^2.\frac{t}{1+x^6}\times \frac{dt}{3x^2}\times (1+x^6)$

$\Rightarrow \int \frac{1}{3}.tdt=\frac{t^2}{6}+c=\frac{\left({\tan }^{-1}{x}^3\right)2}{6}+c$

Hence, this is the answer.