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