Mathematics · 2025
JEE Main · 22 January 2025, Shift 2 · Q20
If ∫ e^x((x sin^-1 x)/(√(1-x^2))+(sin^-1 x)/((1-x^2)^3 / 2)+x/(1-x^2)) d x= g (x)+ C, where C is the constant of integration, then g (1/2) equals:
If $\displaystyle \int \mathrm{e}^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) \mathrm{d} x=\mathrm{g}(x)+\mathrm{C}$, where C is the constant of integration, then $\displaystyle \mathrm{g}\left(\frac{1}{2}\right)$ equals :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{\pi}{6} \sqrt{\frac{e}{3}}$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.