Question

Solve:

$\int \frac{x{\tan }^{-1}{x}^2}{(1+x^4)}dx$

Answer 1

Consider the given integral.

$\int \frac{x{\tan }^{-1}{x}^2}{\left(1+x^4\right)}dx$

Let ${\tan }^{-1}{x}^2=t$

$\frac{dt}{dx}=\frac{2x}{1+x^4}$

$\frac{1}{2}dt=\frac{xdx}{1+x^4}$

Therefore,

$=\frac{1}{2}\frac{t^2}{2}+C=\frac{t^2}{4}+C={\frac{\left[{\tan }^{-1}\left(x^2\right)\right]}{4}}^2+C$

Hence, this is the answer.