Let f ( x )=int frac (2- x 2 ) × e x (sqrt 1+ x )(1- x ) 3 / 2 dx . If f (0)-0 , then f (1/2 ) is equal to :

Mathematics · Indefinite Integration

JEE Main 2026 — 23 January, Morning Shift — Question 2

Let $\mathrm{f}(\mathrm{x})=\int \frac{\left(2-\mathrm{x}^{2}\right) \cdot \mathrm{e}^{\mathrm{x}}}{(\sqrt{1+\mathrm{x}})(1-\mathrm{x})^{3 / 2}} \mathrm{dx}$ . If $\mathrm{f}(0)-0$ , then $\mathrm{f}\left(\frac{1}{2}\right)$ is equal to :

Option A:
$\sqrt{3 \mathrm{e}}-1$
Correct
Option B:
$\sqrt{2 \mathrm{e}}+1$
Option C:
$\sqrt{2 \mathrm{e}}-1$
Option D:
$\sqrt{3 \mathrm{e}}+1$

Answer: A

Step-by-step solution

$\int \mathrm{e}^{\mathrm{x}}\left(\frac{\left(1-\mathrm{x}^{2}\right)+1}{\sqrt{1+\mathrm{x}} \cdot(1-\mathrm{x})^{3 / 2}}\right) \mathrm{dx}$

$\int \mathrm{e}^{\mathrm{x}}\left(\frac{\left(1-\mathrm{x}^{2}\right)}{\sqrt{1+\mathrm{x}} \cdot(1-\mathrm{x})^{3 / 2}}+\frac{1}{\sqrt{1+\mathrm{x}} \cdot(1-\mathrm{x})^{3 / 2}}\right) \mathrm{dx}$

$\int \mathrm{e}^{\mathrm{x}}\left(\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}+\frac{1}{\sqrt{1+\mathrm{x}} \cdot(1-\mathrm{x})^{3 / 2}}\right) \mathrm{dx}$

$=\mathrm{e}^{\mathrm{x}} \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}+\mathrm{C}$

$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}-1$

$\mathrm{f}\left(\frac{1}{2}\right)=\sqrt{3 \mathrm{e}}-1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration
Topic: Methods of Indefinite Integration