The value of int_{0}^{1} frac{tan^{-1} x}{1 + | vedclass

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Question

(B) Let $I = \int_{0}^{1} \frac{	an^{-1} x}{1 + x^2} dx$. Substitute $t = 	an^{-1} x$. Then,$dt = \frac{1}{1 + x^2} dx$. When $x = 0$,$t = 	an^{-1}

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$\pi^2 / 32$

Vedclass · Answer

(B) Let $I = \int_{0}^{1} \frac{	an^{-1} x}{1 + x^2} dx$. Substitute $t = 	an^{-1} x$. Then,$dt = \frac{1}{1 + x^2} dx$. When $x = 0$,$t = 	an^{-1}(0) = 0$. When $x = 1$,$t = 	an^{-1}(1) = \pi / 4$. Therefore,the integral becomes $I = \int_{0}^{\pi / 4} t dt$. Evaluating the integral,we get $I = \left[ \frac{t^2}{2} ight]_{0}^{\pi / 4}$. $I = \frac{(\pi / 4)^2}{2} - 0 = \frac{\pi^2 / 16}{2} = \frac{\pi^2}{32}$.

The value of $\int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} dx$ is

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