The value of the integral $\int\limits_{-1}^3\left(\tan ^{-1} \frac{x}{x^2+1}+\tan ^{-1} \frac{x^2+1}{x}\right) d x$ is equal to

$\pi$ $2 \pi$ $4 \pi$ none of these

$2 \pi$

We have, $\int\limits_{-1}^3\left(\tan ^{-1} \frac{x}{x^2+1}+\tan ^{-1} \frac{x^2+1}{x}\right) d x$ $=\int\limits_{-1}^3\left(\tan ^{-1} \frac{x}{x^2+1}+\cot ^{-1} \frac{x}{x^2+1}\right) d x=\int\limits_{-1}^3 \frac{\pi}{2} d x=2 \pi$