$\int_1^3\left[ an ^{-1}\left(\frac{x}{x^2-1} ight)+ an ^{-1}\left(\frac{x^2-1}{x} ight) ight] d x=$

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

$\begin{aligned} & 	ext { Let } I=\int_1^3\left[	an ^1\left(\frac{x}{x^2-1}ight)+	an ^{-1}\left(\frac{x^2-1}{x}ight)ight] d x \ & \int_1^3\left[	an ^{-1}\left(\frac{x}{x^2-1}ight)+\cot ^{-1}\left(\frac{x}{x^2-1}ight)ight] d x \ & =\int_1^3\left(\frac{\pi}{2}ight) d x=\frac{\pi}{2} \int_1^3 d x=\frac{\pi}{2}[x]_1^3=\pi\end{aligned}$

$\int_1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right] d x=$

Solution