The value of $\int\limits_{-1}^1|\tan^{-1}x| dx$ is:

$\frac{\pi}{2}-\log_e2$

The correct answer is Option (1) → $\frac{\pi}{2}-\log_e2$

$I=\int|\tan^{-1}x| dx$

$⇒I=\int\limits_{x<0}-\tan^{-1}x\,dx+\int\limits_{x≥0}-\tan^{-1}x\,dx$

and,

$\int\limits_{x≥0}\tan^{-1}x\,dx=x\tan^{-1}x-\frac{1}{2}\log|1+x^2|+C$

$∴I=\left\{\begin{matrix}-x\tan^{-1}x+\frac{1}{2}\log|1+x^2|+C,&x<0\\x\tan^{-1}x-\frac{1}{2}\log|1+x^2|+C&x≥0\end{matrix}\right.$

$∴\int\limits_{-1}^1|\tan^{-1}x|=\frac{\pi}{2}-\log_e2$