Evaluate int_{0}^{1} frac{tan ^{-1} x}{1+x^{2 | vedclass

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(C) Let $t = 	an^{-1} x$. Then,$dt = \frac{1}{1+x^2} dx$.Change the limits of integration:When $x = 0$,$t = 	an^{-1}(0) = 0$.When $x = 1$,$t = 	an^

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$\frac{\pi^2}{32}$

Vedclass · Answer

(C) Let $t = 	an^{-1} x$. Then,$dt = \frac{1}{1+x^2} dx$.Change the limits of integration:When $x = 0$,$t = 	an^{-1}(0) = 0$.When $x = 1$,$t = 	an^{-1}(1) = \frac{\pi}{4}$.Substitute these into the integral:$\int_{0}^{1} \frac{	an^{-1} x}{1+x^2} dx = \int_{0}^{\frac{\pi}{4}} t dt$.Evaluate the integral:$\int_{0}^{\frac{\pi}{4}} t dt = \left[ \frac{t^2}{2} ight]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left( \left( \frac{\pi}{4} ight)^2 - 0^2 ight) = \frac{1}{2} \left( \frac{\pi^2}{16} ight) = \frac{\pi^2}{32}$.

Evaluate $\int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x$

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