Q28 / 75 · JEE Main 2025 (2 Apr Shift 2) · Mathematics

$4 \int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3})$ is equal to :

  • A
    $2+\sqrt{2}+\log _e(1+\sqrt{2})$
  • Correct answer
    B
    $2-\sqrt{2}-\log _e(1+\sqrt{2})$
  • C
    $2+\sqrt{2}-\log _e(1+\sqrt{2})$
  • D
    $2-\sqrt{2}+\log _e(1+\sqrt{2})$

Solution

$\begin{aligned} & 4 \int_0^1 \frac{1}{\sqrt{3+\mathrm{x}^2}+\sqrt{1+\mathrm{x}^2}} \mathrm{dx}-3 \ln \sqrt{3} \\ & =4 \int_0^1 \frac{\sqrt{3+\mathrm{x}^2}-\sqrt{1+\mathrm{x}^2}}{\left(3+\mathrm{x}^2\right)-\left(1-\mathrm{x}^2\right)} \mathrm{dx}-\frac{3}{2} \ln 3 \\ & = \\ & 2\left[\left\{\frac{\mathrm{x}}{2} \sqrt{3+\mathrm{x}^2}+\frac{3}{2} \ln \left(\mathrm{x}+\sqrt{3+\mathrm{x}^2}\right)\right\}_0^1\right. \\ & \left.\quad-\left\{\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} \ln \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right\}_0^1\right]-\frac{3}{2} \ln 3\end{aligned}$
$\begin{aligned}= & 2\left[\left\{\frac{1}{2} \sqrt{4}+\frac{3}{2} \ln (1+\sqrt{4})\right\}-\left\{0+\frac{3}{2} \ln \sqrt{3}\right\}\right. \\ & \left.-\left\{\frac{1}{2} \sqrt{2}+\frac{1}{2} \ln (1+\sqrt{2})\right\}+\left\{0+\frac{1}{2}(0)\right\}\right]-\frac{3}{2} \ln 3\end{aligned}$
$\begin{aligned} & =2\left[1+\frac{3}{2} \ln 3-\frac{3}{4} \ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} \ln (1+\sqrt{2})\right]-\frac{3}{2} \ln 3 \\ & =2+3 \ln 3-\frac{3}{2} \ln 3-\sqrt{2}-\ln (1+\sqrt{2})-\frac{3}{2} \ln 3 \\ & =2-\sqrt{2}-\ln (1+\sqrt{2})\end{aligned}$

Hence, the correct answer is option: 'B'.

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