Explanation
\begin{aligned}
& \mathrm{Let}\mathrm{I}=\int_{0}^{1}\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx \\
& =\int_{0}^{1}\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)}dx\quad[\text{on rationalising}] & \mathrm{-} \\
& =\int_{0}^{1}\left[\frac{\sqrt{3+x}-\sqrt{1+x}}{2}\right]dx \\
& =\frac{1}{2}\left\{\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}}\right\}|_{0}^{1} \\
& =\frac{1}{3}\left\{\left[\left(1+3\right)^{\frac{3}{2}}-\left(1+1\right)^{\frac{3}{2}}\right]-\left[\left(0+3\right)^{\frac{3}{2}}-\left(1+0\right)^{\frac{3}{2}}\right]\right\} \\
& =\frac{1}{3}\left\{8-2\sqrt{2}-3\sqrt{3}+1\right\} \\
& =\frac{1}{3}\left\{9-2\sqrt{2}-3\sqrt{3}\right\}=3-\frac{2}{\sqrt{3}}\sqrt{2}-\sqrt{3} \\
& =a+b\sqrt2+c\sqrt3
\end{aligned}
\begin{aligned}
& \Rightarrow a=\frac{3}{0},b=-\frac{2}{3},\mathrm{C}=-1 \\
& \mathrm{So},2a+3b-4c=-2+4=8
\end{aligned}
