Question

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

Answer 1

Consider given the given integration,

Let,

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

Put,

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

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

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

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

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

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

Hence, this is the answer.