Question

Evaluate $\int {\tan }^{-1}xdx$

Answer 1

Let $I=\int {\tan }^{-1}xdx$

Applying by parts,

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

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

$=x{\tan }^{-1}x-I_1$

Let $I_1=\int \frac{x}{1+x^2}dx$, then, put $1+x^2=t$ and $2xdx=dt$.

$I_1=\frac{1}{2}\int \frac{dt}{t}$

$=\frac{1}{2}\log t+c$

$=\frac{1}{2}\log \left(1+x^2\right)+c$

Therefore,

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