Question

$\int \frac{x^2{\tan }^{-1}{x}^3}{1+x^6}dx$ is equal to

Answer 1

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

$put{\tan }^{-1}{x}^3\sigma$

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

$\int tdt$

$\frac{1}{3}\times \frac{t^2}{2}tc=\frac{({\tan }^{-}{x}^3{)}^2}{6}tc$