The value of the integral int -1 2 log e (x+sqrt x 2 +1 ) d x is :

√(2)-√(5)+log e (frac 9+4 √(5) 1+√(2) )

√(5)-√(2)+log e (frac 9+4 √(5) 1+√(2) )

√(5)-√(2)+log e (frac 7+4 √(5) 1+√(2) )

√(2)-√(5)+log e (frac 7+4 √(5) 1+√(2) )

Mathematics · Definite Integration

JEE Main 2024 — 9 April, Shift 2 — Question 18

The value of the integral $\int_{-1}^{2} \log _{e}\left(x+\sqrt{x^{2}+1}\right) d x$ is :

Option A:
$\sqrt{5}-\sqrt{2}+\log _{\mathrm{e}}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$
Option B:
$\sqrt{2}-\sqrt{5}+\log _{\mathrm{e}}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$
Correct
Option C:
$\sqrt{5}-\sqrt{2}+\log _{e}\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$
Option D:
$\sqrt{2}-\sqrt{5}+\log _{\mathrm{e}}\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$

$\quad I=\int_{-1}^{2} 1 . \log _{e}\left(x+\sqrt{x^{2}+1}\right) d x$

$=x \log _{e}\left(x+\sqrt{x^{2}+1}\right)-\int_{-1}^{2}\left(\frac{1+\frac{x}{\sqrt{x^{2}+1}}}{x+\sqrt{x^{2}+1}}\right) d x$

$=x \log _{e}\left(x+\sqrt{x^{2}+1}\right)-\int_{-1}^{2} \frac{x}{\sqrt{x^{2}+1}} d x$

$=x \log _{e}\left(x+\sqrt{x^{2}+1}\right)-\left.\sqrt{x^{2}+1}\right|_{-1} ^{2}$

$=\left(2 \log _{e}(2+\sqrt{5})-\sqrt{5}\right)$ $-\left(-\log _{e}(-1+\sqrt{2})-\sqrt{2}\right)$

$=\log _{e}(2+\sqrt{5})^{2}-\sqrt{5}+\log _{e}(\sqrt{2}-1)+\sqrt{2}$

=\sqrt{2}-\sqrt{5}+{{\log }_{e}}\left( \frac{{{\left( 2+\sqrt{5} \right)}^{2}}}{\sqrt{2}+1} \right)

=\sqrt{2}+\sqrt{5}+{{\log }_{e}}\left( \frac{9+4\sqrt{5}}{\sqrt{2}+1} \right)

Answer key and solution verified before publishing.

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