JEE Main 2024 — 27 January, Shift 1 — Question 9

$\int_{0}^{1} \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_{0}^{1} \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x$

$\frac{1}{2}\left[\int_{0}^{1} \sqrt{3+x} d x-\int_{0}^{1}(\sqrt{1+x}) \mathrm{dx}\right]$

$\frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_{0}^{1}$

$\frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right]$

$\frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1]$

$=3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3}$

$a=3, b=-\frac{2}{3}, c=-1$

$2 a+3 b-4 c=6-2+4=8$