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