$$\begin{aligned} &\begin{aligned} & \because \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}\right)=\frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^2\right)^{3 / 2}}+\frac{\mathrm{x}}{1-\mathrm{x}^2} \\ & \Rightarrow \int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}+\frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^2\right)^{3 / 2}}+\frac{\mathrm{x}}{1-\mathrm{x}^2}\right) \mathrm{dx} \\ & =\mathrm{e}^{\mathrm{x}} \cdot \frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}+\mathrm{c}=\mathrm{g}(\mathrm{x})+\mathrm{C} \end{aligned}\\ &\text { Note : assuming } g(x)=\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} x}{\sqrt{1-\mathrm{x}^2}}\\ &g(1 / 2)=\frac{\mathrm{e}^{1 / 2}}{2} \cdot \frac{\frac{\pi}{6} \times 2}{\sqrt{3}}=\frac{\pi}{6} \sqrt{\frac{\mathrm{e}}{3}} \end{aligned}$$

Comment : In this question we will not get a unique function $\mathrm{g}(\mathrm{x})$, but in order to match the answer we will have to assume $g(x)=\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}$.