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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Indefinite Integration

If ∫ e^x( x ^ -1 x 1-x^2 + ^ -1 x (1-x^2)^ 3 / 2 +(x)/(1-x^2)) d...

If ∫ex(xsin⁡−1x1−x2+sin⁡−1x(1−x2)3/2+x1−x2)dx=g(x)+C\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 g(12)g\left(\frac{1}{2}\right) equals :

Options

  1. Aπ4e3\frac{\pi}{4} \sqrt{\frac{e}{3}}
  2. Bπ6e3\frac{\pi}{6} \sqrt{\frac{e}{3}}
  3. Cπ4e2\frac{\pi}{4} \sqrt{\frac{e}{2}}
  4. Dπ6e2\frac{\pi}{6} \sqrt{\frac{e}{2}}

Correct answer

B. π6e3\frac{\pi}{6} \sqrt{\frac{e}{3}}

Step-by-step solution

ddx(x⋅sin⁡−1x1−x2)−sin⁡−1x⋅(1⋅1−x2+x⋅2x21−x21−x2)=x1−x2⋅11−x2=sin⁡−1x(1−x2)3/2+x1−x2\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} Hence, I=∫ex(f(x)+f′(x))dx=ex⋅f(x)+CI=ex⋅x⋅sin⁡−1x1−x2+C=g(x)+C⇒g(x)=xexsin⁡−1x1−x2 and  g(1/2)=π6e3\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{\text{ and } } g(1 / 2)=\frac{\pi}{6} \sqrt{\frac{e}{3}} \end{aligned}$
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