If e^x(x ^-1 x√1-x^2+ ^-1 x(1-x^2)^3 / 2+(x)/(1-x^2)) d x=g(x)+C, where C is the constant …

Question

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 :

Accepted Answer

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

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 $g(x)=\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}$.

Solutions · 1