int_{-1}^3left(cot ^{-1}left(frac{x}{x^2+1}ri | vedclass

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(D) Let $I = \int_{-1}^3 \left(\cot^{-1}\left(\frac{x}{x^2+1}ight) + \cot^{-1}\left(\frac{x^2+1}{x}ight)ight) dx$. Since $\cot^{-1}(u) = 	an^{-

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$2\pi$

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(D) Let $I = \int_{-1}^3 \left(\cot^{-1}\left(\frac{x}{x^2+1}ight) + \cot^{-1}\left(\frac{x^2+1}{x}ight)ight) dx$. Since $\cot^{-1}(u) = 	an^{-1}(\frac{1}{u})$ for $u > 0$,we observe the term $\cot^{-1}(\frac{x}{x^2+1})$. For $x > 0$,$\cot^{-1}(\frac{x}{x^2+1}) = 	an^{-1}(\frac{x^2+1}{x})$. However,the identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ holds for all $u \in \mathbb{R}$. Thus,the integrand simplifies to $\frac{\pi}{2}$ for all $x 
eq 0$. $I = \int_{-1}^3 \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^3$. $I = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

$\int_{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) d x=$

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