Question

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

Answer 1

We have,

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

Let

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

$\frac{dt}{dx}=\frac{1}{1+(x^3{)}^2}\times 3x^2$

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

Therefore,

$I=\frac{1}{3}\int tdt$

$I=\frac{1}{3}\frac{t^2}{2}+C$

On putting the value of $t$, we get

$I=\frac{1}{6}({\tan }^{-1}{x}^3{)}^2+C$

Hence, this is the answer.