int_{0}^{1} frac{d}{dx} left[ sin^{-1} left( | vedclass

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(C) Let $I = \int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$. By the Fundamental Theorem of Calculus,$\int_{a}^

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$\pi/2$

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(C) Let $I = \int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$. By the Fundamental Theorem of Calculus,$\int_{a}^{b} f'(x) dx = f(b) - f(a)$. Here,$f(x) = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$. So,$I = \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right]_{0}^{1}$. Evaluating at the limits: $I = \sin^{-1} \left( \frac{2(1)}{1+(1)^2} \right) - \sin^{-1} \left( \frac{2(0)}{1+(0)^2} \right)$. $I = \sin^{-1} \left( \frac{2}{2} \right) - \sin^{-1} (0)$. $I = \sin^{-1} (1) - \sin^{-1} (0)$. Since $\sin^{-1} (1) = \frac{\pi}{2}$ and $\sin^{-1} (0) = 0$,$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

$\int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$ is equal to

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