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
Answer & explanation
Correct answer: option 2
We have,
$\int\limits_{-1}^3\left(\tan ^{-1} \frac{x}{x^2+1}+\tan ^{-1} \frac{x^2+1}{x}\right) d x$
$=\int\limits_{-1}^3\left(\tan ^{-1} \frac{x}{x^2+1}+\cot ^{-1} \frac{x}{x^2+1}\right) d x=\int\limits_{-1}^3 \frac{\pi}{2} d x=2 \pi$