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JEE Main 09-Apr-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 09-Apr-2024 Paper - Shift 24 Marks
MCQ
The value of the integral $\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ is :
  • A
    $\sqrt{5}-\sqrt{2}+\log _{ c }\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$
  • B
    $\sqrt{2}-\sqrt{5}+\log _c\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$
  • C
    $\sqrt{5}-\sqrt{2}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$
  • D
    $\sqrt{2}-\sqrt{5}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$
✓

Answer

$I=\int_{-1}^2 1_1 \log _e\left(x+\sqrt{x^2+1}\right) d x$
$=x \log _e\left(x+\sqrt{x^2+1}\right)-\int_{-1}^2\left(\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\right) d x$
$=x \log _e\left(x+\sqrt{x^2+1}\right)-\int_{-1}^2 \frac{x}{\sqrt{x^2+1}} d x$
$=x \log _e\left(x+\sqrt{x^2+1}\right)-\left.\sqrt{x^2+1}\right|_{-1} ^2$
$=\left(2 \log _e(2+\sqrt{5})-\sqrt{5}\right)$
$-\left(-\log _e(-1+\sqrt{2})-\sqrt{2}\right)$
$=\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2}$
$=\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2}$
$=\sqrt{2}-\sqrt{5}+\log _{ e }\left(\frac{(2+\sqrt{5})^2}{\sqrt{2+1}}\right)$
$=\sqrt{2}-\sqrt{5}+\log _{ c }\left(\frac{9+4 \sqrt{5}}{\sqrt{2+1}}\right)$

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