int_1^3 left[ tan^{-1} left( frac{x}{x^2-1} r | vedclass

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(A) We know that $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.Given the integral $I = \int_1^3 \left[ 	an^{-1} \left( \fr

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$\pi$

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(A) We know that $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.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)$,the expression becomes $	an^{-1} \left( \frac{x}{x^2-1} ight) + \cot^{-1} \left( \frac{x}{x^2-1} ight)$.Thus,the integrand simplifies to $\frac{\pi}{2}$.Therefore,$I = \int_1^3 \frac{\pi}{2} dx = \frac{\pi}{2} [x]_1^3 = \frac{\pi}{2} (3-1) = \frac{\pi}{2} 	imes 2 = \pi$.

$\int_1^3 \left[ \tan^{-1} \left( \frac{x}{x^2-1} \right) + \tan^{-1} \left( \frac{x^2-1}{x} \right) \right] dx =$

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