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