Correct option is (4) 8
\(\int_{0}^{1} \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_{0}^{1} \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x\)
\(\frac{1}{2}\left[\int_{0}^{1} \sqrt{3+x} d x-\int_{0}^{1}(\sqrt{1+x}) \mathrm{dx}\right]\)
\(\frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_{0}^{1}\)
\(\frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right]\)
\(\frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1]\)
\(=3-\sqrt{3}-\frac{2}{3} \sqrt{2}=\mathrm{a}+\mathrm{b} \sqrt{2}+\mathrm{c} \sqrt{3}\)
\(\mathrm{a}=3, \mathrm{~b}=-\frac{2}{3}, \mathrm{c}=-1\)
\(2 \mathrm{a}+3 \mathrm{~b}-4 \mathrm{c}=6-2+4=8\)
