$\int_{-1}^{3}\left[ an ^{-1}\left(\frac{x}{x^{2}+1} ight)+ an ^{-1}\left(\frac{x^{2}+1}{x} ight) ight] d x=$

Question

$\int_{-1}^{3}\left[	an ^{-1}\left(\frac{x}{x^{2}+1}ight)+	an ^{-1}\left(\frac{x^{2}+1}{x}ight)ight] d x=$

Aicharya · Accepted Answer

$\begin{aligned} I &=\int_{-1}^{3}\left[	an ^{-1}\left(\frac{x}{x^{2}+1}ight)+	an ^{-1}\left(\frac{x^{2}+1}{x}ight)ight] d x \ &=\int_{-1}^{3}\left[	an ^{-1}\left(\frac{x}{x^{2}+1}ight)+\cot ^{-1}\left(\frac{x}{x^{2}+1}ight)ight] d x \ &=\int_{-1}^{3} \frac{\pi}{2} d x=\frac{\pi}{2}[x]_{-1}^{3}=\frac{4 \pi}{2} \ &=2 \pi \end{aligned}$

$\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right]…

Solution