Question

$\int {\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)dx \\ =\int 3{\tan }^{-1}\left(x\right)dx \\ =3\int \left[{\tan }^{-1}\left(x\right)\times 1\right]dx \\ =3\left[{\tan }^{-1}x\times x-\int \frac{1}{1+x^2}\times xdx\right] \\ =3x{\tan }^{-1}x-3\int \frac{x}{1+x^2}dx \\ \mathrm{let}1+x^2=t \\ \Rightarrow 2xdx=dt \\ \mathrm{Then}, \\ I=3x{\tan }^{-1}x-\frac{3}{2}\int \frac{dt}{t} \\ =3x{\tan }^{-1}x-\frac{3}{2}\log \left|t\right|+C \\ =3x{\tan }^{-1}x-\frac{3}{2}\log \left|1+x^2\right|+C \end{gathered}$$