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

$2 \pi$

$I=\int\limits_{-1}^3\left[\tan ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x}{x^2+1}\right)\right] dx=\int\limits_{-1}^3 \frac{\pi}{2} dx=2 \pi$

Hence (2) is the correct answer.