JEE Main 2024 — Definite Integration Question with Solution

\begin{aligned} & \int_\limits0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_\limits0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\ & \frac{1}{2}\left[\int_\limits0^1 \sqrt{3+x} d x-\int_\limits0^1(\sqrt{1+x}) d x\right] \end{aligned}

$$\begin{array}{rl}& \frac{1}{2}{\left[2\frac{(3+x{)}^{\frac{3}{2}}}{3}-\frac{2(1+x{)}^{\frac{3}{2}}}{3}\right]}_0^1\\ & \frac{1}{2}\left[\frac{2}{3}(8-3\sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right]\\ & \frac{1}{3}[8-3\sqrt{3}-2\sqrt{2}+1]\\ & =3-\sqrt{3}-\frac{2}{3}\sqrt{2}=a+b\sqrt{2}+c\sqrt{3}\\ & a=3,b=-\frac{2}{3},c=-1\\ & 2a+3b-4c=6-2+4=8\end{array}$$