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

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(B) Let $I = \int_{-1}^3 \left(	an^{-1}\left(\frac{x}{x^2+1}ight) + 	an^{-1}\left(\frac{x^2+1}{x}ight)ight) dx$. Using the property $	an^{-1}

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$\pi$

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(B) Let $I = \int_{-1}^3 \left(	an^{-1}\left(\frac{x}{x^2+1}ight) + 	an^{-1}\left(\frac{x^2+1}{x}ight)ight) dx$. Using the property $	an^{-1}(u) + 	an^{-1}(1/u) = \frac{\pi}{2}$ for $u > 0$,we observe that the expression inside the integral is $\frac{\pi}{2}$ whenever $x > 0$. However,for $x Using $	an^{-1}(u) + 	an^{-1}(1/u) = -\frac{\pi}{2}$ for $u $I = \int_{-1}^0 (-\frac{\pi}{2}) dx + \int_0^3 (\frac{\pi}{2}) dx$. $I = -\frac{\pi}{2} [x]_{-1}^0 + \frac{\pi}{2} [x]_0^3$. $I = -\frac{\pi}{2} (0 - (-1)) + \frac{\pi}{2} (3 - 0)$. $I = -\frac{\pi}{2} (1) + \frac{\pi}{2} (3) = -\frac{\pi}{2} + \frac{3\pi}{2} = \frac{2\pi}{2} = \pi$.

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

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