If int e^xleft(frac{x sin ^{-1} x}{sqrt{1-x^2 | vedclass

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(C) We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.Let $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.Then $f'(x) = \frac{d}{dx} \left( \frac{x

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$\frac{\pi}{6} \sqrt{\frac{e}{3}}$

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(C) We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.Let $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.Then $f'(x) = \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) = \frac{\sqrt{1-x^2} \cdot \frac{d}{dx}(x \sin^{-1} x) - x \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1-x^2})}{1-x^2}$$= \frac{\sqrt{1-x^2} \left( \sin^{-1} x + \frac{x}{\sqrt{1-x^2}} \right) - x \sin^{-1} x \left( \frac{-x}{\sqrt{1-x^2}} \right)}{1-x^2}$$= \frac{\sqrt{1-x^2} \sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2} = \frac{(1-x^2) \sin^{-1} x + x \sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)^{3/2}} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$.Thus,the integral is $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + C$.So,$g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1-x^2}}$.Evaluating at $x = \frac{1}{2}$: $g\left(\frac{1}{2}\right) = \frac{\frac{1}{2} e^{1/2} \sin^{-1}(1/2)}{\sqrt{1-(1/2)^2}} = \frac{\frac{1}{2} \sqrt{e} \cdot \frac{\pi}{6}}{\sqrt{3/4}} = \frac{\frac{\pi \sqrt{e}}{12}}{\frac{\sqrt{3}}{2}} = \frac{\pi \sqrt{e}}{6 \sqrt{3}} = \frac{\pi}{6} \sqrt{\frac{e}{3}}$.

If $\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C$ where $C$ is the constant of integration,then $g \left(\frac{1}{2}\right)$ equals :

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