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

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \frac{d}{d x}\left(\frac{x \cdot \sin ^{-1} x}{\sqrt{1-x^2}}\right)-\sin ^{-1} x \cdot\left(\frac{1 \cdot \sqrt{1-x^2}+\frac{x \cdot 2 x}{2 \sqrt{1-x^2}}}{1-x^2}\right) \\ & =\frac{x}{\sqrt{1-x^2}} \cdot \frac{1}{\sqrt{1-x^2}} \\ & =\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}
\end{aligned}$
$\begin{aligned}
& \text { Hence, } I=\int e^x\left(f(x)+f^{\prime}(x)\right) d x \\ & =e^x \cdot f(x)+C \\ & I=e^x \cdot \frac{x \cdot \sin ^{-1} x}{\sqrt{1-x^2}}+C=g(x)+C \\ & \Rightarrow g(x)=\frac{x e^x \sin ^{-1} x}{\sqrt{1-x^2}} \text { and } g(1 / 2)=\frac{\pi}{6} \sqrt{\frac{e}{3}}
\end{aligned}$

If $\int \mathrm{e}^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 /…

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