intlimits_0^{sqrt{3}} frac{1}{2} frac{d}{dx} | vedclass

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(B) Let $f(x) = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$.We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = \begin{cases} 2	an^{-1}x & 	ext

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$-\frac{\pi}{6}$

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(B) Let $f(x) = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$.We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = \begin{cases} 2	an^{-1}x & 	ext{if } |x|  1 \end{cases}$.The integral is $I = \int_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} [f(x)] dx = \frac{1}{2} [f(x)]_0^{\sqrt{3}} = \frac{1}{2} [f(\sqrt{3}) - f(0)]$.For $x = \sqrt{3} > 1$,$f(\sqrt{3}) = 2	an^{-1}(\sqrt{3}) - \pi = 2(\frac{\pi}{3}) - \pi = \frac{2\pi}{3} - \pi = -\frac{\pi}{3}$.For $x = 0$,$f(0) = 2	an^{-1}(0) = 0$.Thus,$I = \frac{1}{2} [-\frac{\pi}{3} - 0] = -\frac{\pi}{6}$.

$\int\limits_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} \left( \tan^{-1} \frac{2x}{1-x^2} \right) dx$ equals

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