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

(C) $\frac{\pi}{6} \sqrt{\frac{e}{3}}$

(A) $\frac{\pi}{6} \sqrt{\frac{e}{2}}$

(B) $\frac{\pi}{4} \sqrt{\frac{e}{2}}$

(D) $\frac{\pi}{4} \sqrt{\frac{e}{3}}$

Indefinite Integrals — JEE Main · 2025 · 22 Jan 2025 · Shift 2 Question with Solution

\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) dx = g (x) + C

, where C is the constant of integration, then

g (\frac{1}{2})

equals :

Answer: (C)

Observe that

\frac{d}{d x} (\frac{x sin ^{- 1} x}{1 - x ^{2}}) = \frac{sin ^{- 1} x}{1 - x ^{2}} + \frac{x}{1 - x ^{2}} + \frac{x ^{2} sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} .

Writing

x^{2} = 1 - (1 - x^{2})

, the last term splits as

\frac{x ^{2} sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} = - \frac{sin ^{- 1} x}{1 - x ^{2}} + \frac{sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} .

So the derivative simplifies to

\frac{d}{d x} (\frac{x sin ^{- 1} x}{1 - x ^{2}}) = \frac{x}{1 - x ^{2}} + \frac{sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} .

Adding

\frac{x sin ^{- 1} x}{1 - x ^{2}}

to both sides gives exactly the bracketed integrand:

\frac{x sin ^{- 1} x}{1 - x ^{2}} + \frac{sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}} = f (x) + f^{'} (x), f (x) = \frac{x sin ^{- 1} x}{1 - x ^{2}} .

Since

\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C

, taking this natural antiderivative (its constant absorbed into C) gives

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

x = \frac{1}{2}

g (\frac{1}{2}) = \frac{\frac{1}{2} \cdot e ^{1/2} \cdot \frac{π}{6}}{3 /2} = \frac{π}{6} \cdot \frac{e}{3} = \frac{π}{6} \frac{e}{3} .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option