If ∫ 0 1 1 3 + x + 1 + x d x = a + b 2 + c 3 , where a , b , c are rational numbers, then 2 a + 3 b - 4 c is…

Let, $I={\int }_0^1\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx$

$\Rightarrow I={\int }_0^1\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)}dx$

$\Rightarrow I=\frac{1}{2}\left[{\int }_0^1\sqrt{3+x}dx-{\int }_0^1\left(\sqrt{1+x}\right)dx\right]$

$\Rightarrow I=\frac{1}{2}{\left[\frac{2(3+x{)}^{\frac{3}{2}}}{3}-\frac{2(1+x{)}^{\frac{3}{2}}}{3}\right]}_0^1$

$\Rightarrow I=\frac{1}{2}\left\{\left[\frac{2(3+1{)}^{\frac{3}{2}}}{3}-\frac{2(1+1{)}^{\frac{3}{2}}}{3}\right]-\left[\frac{2(3{)}^{\frac{3}{2}}}{3}-\frac{2(1{)}^{\frac{3}{2}}}{3}\right]\right\}$

$\Rightarrow I=\frac{1}{2}\left\{\frac{2\times 8}{3}-\frac{4\sqrt{2}}{3}-\frac{2\times 3\sqrt{3}}{3}+\frac{2}{3}\right\}$

$\Rightarrow I=\frac{1}{2}\left\{6-\frac{4\sqrt{2}}{3}-2\sqrt{3}\right\}$

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

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

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