int_{-1}^{3} left[ tan^{-1} left( frac{x}{x^{ | vedclass

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(B) We know that $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$. Given the integral $I = \int_{-1}^{3} \left[ 	an^{-1} \left( \frac{x}

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$2 \pi$

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(B) We know that $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$. Given the integral $I = \int_{-1}^{3} \left[ 	an^{-1} \left( \frac{x}{x^{2}+1} ight) + 	an^{-1} \left( \frac{x^{2}+1}{x} ight) ight] dx$. Since $	an^{-1} \left( \frac{x^{2}+1}{x} ight) = \cot^{-1} \left( \frac{x}{x^{2}+1} ight)$ for $x > 0$,the expression inside the integral becomes $	an^{-1} \left( \frac{x}{x^{2}+1} ight) + \cot^{-1} \left( \frac{x}{x^{2}+1} ight) = \frac{\pi}{2}$. Thus,$I = \int_{-1}^{3} \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2 \pi$.

$\int_{-1}^{3} \left[ \tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \tan^{-1} \left( \frac{x^{2}+1}{x} \right) \right] dx =$

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