int_{-1}^{1} x tan^{-1} x , dx equals | vedclass

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(A) Let $I = \int_{-1}^{1} x 	an^{-1} x \, dx$. Since $f(x) = x 	an^{-1} x$ is an even function (because $f(-x) = (-x) 	an^{-1}(-x) = (-x)(-	an^{-

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$\frac{\pi}{2} - 1$

Vedclass · Answer

(A) Let $I = \int_{-1}^{1} x 	an^{-1} x \, dx$. Since $f(x) = x 	an^{-1} x$ is an even function (because $f(-x) = (-x) 	an^{-1}(-x) = (-x)(-	an^{-1} x) = x 	an^{-1} x = f(x)$),we can write: $I = 2 \int_{0}^{1} x 	an^{-1} x \, dx$. Using integration by parts,let $u = 	an^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$. $I = 2 \left[ \frac{x^2}{2} 	an^{-1} x \Big|_0^1 - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx ight]$ $I = \left[ x^2 	an^{-1} x ight]_0^1 - \int_0^1 \frac{x^2}{1+x^2} \, dx$ $I = (1^2 	an^{-1} 1 - 0) - \int_0^1 \frac{x^2+1-1}{1+x^2} \, dx$ $I = \frac{\pi}{4} - \int_0^1 \left( 1 - \frac{1}{1+x^2} ight) \, dx$ $I = \frac{\pi}{4} - [x - 	an^{-1} x]_0^1$ $I = \frac{\pi}{4} - (1 - 	an^{-1} 1) = \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1$.

$\int_{-1}^{1} x \tan^{-1} x \, dx$ equals

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