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$\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x=$
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The correct answer is:
$(2 \pi)$
$\begin{array}{rlr}\text { Let } \mathrm{I} & =\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \\ & =\int_{-1}^3\left(\tan ^{-1}\left(\frac{x^2+1}{x}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \\ & =\int_{-1}^3 \frac{\pi}{2} \mathrm{~d} x \quad \ldots\left[\because \cot ^{-1}(x)=\tan ^{-1}\left(\frac{1}{x}\right)\right] \\ & =\frac{\pi}{2}[x]_{-1}^3 & \\ & =\frac{\pi}{2}(4) & \end{array}$
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