JEE Main 2025 — Indefinite Integration Question with Solution
JEE Main 2025 (22 Jan Shift 2)
Question
If , where C is the constant of integration, then equals :
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
$\begin{aligned}
& \frac{d}{d x}\left(\frac{x \cdot \sin ^{-1} x}{\sqrt{1-x^2}}\right)-\sin ^{-1} x \cdot\left(\frac{1 \cdot \sqrt{1-x^2}+\frac{x \cdot 2 x}{2 \sqrt{1-x^2}}}{1-x^2}\right) \\
& =\frac{x}{\sqrt{1-x^2}} \cdot \frac{1}{\sqrt{1-x^2}} \\
& =\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}
\end{aligned}$
$\begin{aligned}
& \text { Hence, } I=\int e^x\left(f(x)+f^{\prime}(x)\right) d x \\
& =e^x \cdot f(x)+C \\
& I=e^x \cdot \frac{x \cdot \sin ^{-1} x}{\sqrt{1-x^2}}+C=g(x)+C \\
& \Rightarrow g(x)=\frac{x e^x \sin ^{-1} x}{\sqrt{1-x^2}} \text { and } g(1 / 2)=\frac{\pi}{6} \sqrt{\frac{e}{3}}
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Indefinite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.