Question

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

Answer 1

$$\begin{gathered} \int \frac{x{\tan }^{-1}{x}^2}{1+x^4}dx \\ \mathrm{Let}{\tan }^{-1}{x}^2=t \\ \Rightarrow \frac{1}{1+{\left(x^2\right)}^2}\times 2x=\frac{dt}{dx} \\ \Rightarrow \frac{xdx}{1+x^4}=\frac{dt}{2} \\ \mathrm{Now},\int \frac{x{\tan }^{-1}{x}^2}{1+x^4}dx \\ =\frac{1}{2}\int t.dt \\ =\frac{1}{2}\times \frac{t^2}{2}+\mathrm{C} \\ =\frac{{\left({\tan }^{-1}{x}^2\right)}^2}{4}+\mathrm{C} \end{gathered}$$