int_1^2 left( tan ^{-1}left(frac{x}{x^2+1}rig | vedclass

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(D) We know that for $y > 0$,$	an ^{-1}(y) + 	an ^{-1}(1/y) = \frac{\pi}{2}$.Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.Therefore,the i

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$\frac{\pi}{2}$

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(D) We know that for $y > 0$,$	an ^{-1}(y) + 	an ^{-1}(1/y) = \frac{\pi}{2}$.Let $y = \frac{x}{x^2+1}$. Since $x \in [1, 2]$,$y > 0$.Therefore,the integrand simplifies as follows:$	an ^{-1}\left(\frac{x}{x^2+1}ight) + 	an ^{-1}\left(\frac{x^2+1}{x}ight) = \frac{\pi}{2}$.Now,evaluate the integral:$I = \int_1^2 \frac{\pi}{2} d x$$I = \frac{\pi}{2} [x]_1^2$$I = \frac{\pi}{2} (2 - 1) = \frac{\pi}{2}$.

$\int_1^2 \left( \tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right) \right) d x =$

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