JEE Main 2025 Jan 22 Shift 2, Mathematics Q11: Integration by Parts

The integrand has the form $e^x(f(x)+f^{\prime }(x))$ for which the standard result is $\int e^x(f+f^{\prime })dx=e^{x}f(x)+C$.

Let $f(x)=\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}$.

Differentiating $f$ using the quotient and product rules, $f^{\prime }(x)=\frac{{\sin }^{-1}x}{(1-x^2{)}^{3/2}}+\frac{x}{1-x^2}$, which matches the remaining two terms inside the bracket.

Therefore $g(x)=e^x\cdot \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}$.

At $x=\frac{1}{2}$: ${\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$, $1-x^2=\frac{3}{4}$, so $\sqrt{1-x^2}=\frac{\sqrt{3}}{2}$.

$g\left(\frac{1}{2}\right)=e^{1/2}\cdot \frac{\frac{1}{2}\cdot \frac{\pi }{6}}{\frac{\sqrt{3}}{2}}=\sqrt{e}\cdot \frac{\pi }{6\sqrt{3}}=\frac{\pi }{6}\sqrt{\frac{e}{3}}$.