Q39 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let $f(x)=\int \frac{\left(2-x^{2}\right) \cdot \mathrm{e}^{x}}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} x$. If $f(0)=0$, then $f\left(\frac{1}{2}\right)$ is equal to:
- Correct answerA$\sqrt{3 \mathrm{e}}-1$
- B$\sqrt{3 \mathrm{e}}+1$
- C$\sqrt{2 \mathrm{e}}+1$
- D$\sqrt{2 \mathrm{e}}-1$
Solution
Write $2 - x^2 = (1-x)(1+x) + 1$, so the integrand splits as $e^x\left(\sqrt{\frac{1+x}{1-x}} + \frac{1}{\sqrt{1+x}(1-x)^{3/2}}\right)$.
Let $g(x) = \sqrt{\frac{1+x}{1-x}}$. Then $g'(x) = \frac{1}{\sqrt{1+x}(1-x)^{3/2}}$.
Using $\int e^x(g + g')dx = e^x g + C$: $f(x) = e^x\sqrt{\frac{1+x}{1-x}} + C$.
$f(0) = 1 + C = 0 \Rightarrow C = -1$.
$f(1/2) = e^{1/2}\sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{e}\cdot\sqrt{3} - 1 = \sqrt{3e} - 1$.
Let $g(x) = \sqrt{\frac{1+x}{1-x}}$. Then $g'(x) = \frac{1}{\sqrt{1+x}(1-x)^{3/2}}$.
Using $\int e^x(g + g')dx = e^x g + C$: $f(x) = e^x\sqrt{\frac{1+x}{1-x}} + C$.
$f(0) = 1 + C = 0 \Rightarrow C = -1$.
$f(1/2) = e^{1/2}\sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{e}\cdot\sqrt{3} - 1 = \sqrt{3e} - 1$.
Hence, the correct answer is option: 'A'.