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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Definite Integration

4 ∫_0^1( 1 3+x^2 + 1+x^2 ) d x-3 _e( 3 ) is equal to :

4∫01(13+x2+1+x2)dx−3log⁡e(3)4 \int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3}) is equal to :

Options

  1. A2+2+log⁡e(1+2)2+\sqrt{2}+\log _e(1+\sqrt{2})
  2. B2−2−log⁡e(1+2)2-\sqrt{2}-\log _e(1+\sqrt{2})
  3. C2+2−log⁡e(1+2)2+\sqrt{2}-\log _e(1+\sqrt{2})
  4. D2−2+log⁡e(1+2)2-\sqrt{2}+\log _e(1+\sqrt{2})

Correct answer

B. 2−2−log⁡e(1+2)2-\sqrt{2}-\log _e(1+\sqrt{2})

Step-by-step solution

4∫0113+x2+1+x2dx−3ln⁡3=4∫013+x2−1+x2(3+x2)−(1−x2)dx−32ln⁡3=2[{x23+x2+32ln⁡(x+3+x2)}01−{x21+x2+12ln⁡(x+1+x2)}01]−32ln⁡3\begin{aligned} & 4 \int_0^1 \frac{1}{\sqrt{3+\mathrm{x}^2}+\sqrt{1+\mathrm{x}^2}} \mathrm{dx}-3 \ln \sqrt{3} \\ & =4 \int_0^1 \frac{\sqrt{3+\mathrm{x}^2}-\sqrt{1+\mathrm{x}^2}}{\left(3+\mathrm{x}^2\right)-\left(1-\mathrm{x}^2\right)} \mathrm{dx}-\frac{3}{2} \ln 3 \\ & = \\ & 2\left[\left\{\frac{\mathrm{x}}{2} \sqrt{3+\mathrm{x}^2}+\frac{3}{2} \ln \left(\mathrm{x}+\sqrt{3+\mathrm{x}^2}\right)\right\}_0^1\right. \\ & \left.\quad-\left\{\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} \ln \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right\}_0^1\right]-\frac{3}{2} \ln 3\end{aligned}
=2[{124+32ln⁡(1+4)}−{0+32ln⁡3}−{122+12ln⁡(1+2)}+{0+12(0)}]−32ln⁡3\begin{aligned}= & 2\left[\left\{\frac{1}{2} \sqrt{4}+\frac{3}{2} \ln (1+\sqrt{4})\right\}-\left\{0+\frac{3}{2} \ln \sqrt{3}\right\}\right. \\ & \left.-\left\{\frac{1}{2} \sqrt{2}+\frac{1}{2} \ln (1+\sqrt{2})\right\}+\left\{0+\frac{1}{2}(0)\right\}\right]-\frac{3}{2} \ln 3\end{aligned}
=2[1+32ln⁡3−34ln⁡3−12−12ln⁡(1+2)]−32ln⁡3=2+3ln⁡3−32ln⁡3−2−ln⁡(1+2)−32ln⁡3=2−2−ln⁡(1+2)\begin{aligned} & =2\left[1+\frac{3}{2} \ln 3-\frac{3}{4} \ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} \ln (1+\sqrt{2})\right]-\frac{3}{2} \ln 3 \\ & =2+3 \ln 3-\frac{3}{2} \ln 3-\sqrt{2}-\ln (1+\sqrt{2})-\frac{3}{2} \ln 3 \\ & =2-\sqrt{2}-\ln (1+\sqrt{2})\end{aligned}
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