Question

Write a value of $\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx$.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{{\left({\tan }^{-1}x\right)}^3}{1+x^2}dx \\ \mathrm{Let}{\tan }^{-1}x=t \\ \Rightarrow \frac{1}{1+x^2}dx=dt \\ \therefore I=\int t^3.dt \\ =\frac{t^4}{4}+C \\ =\frac{{\left({\tan }^{-1}x\right)}^4}{4}+C\left(\because t={\tan }^{-1}x\right) \end{gathered}$$