If ∫ _ 0 ^ 1 1 3 + x + 1 + x dx = a + b 2 + c 3 , where a, b, c are...

Let, $I = {∫}_{0}^{1}\dfrac{1}{\sqrt{3 + x} + \sqrt{1 + x}}dx$

$⇒ I = {∫}_{0}^{1}\dfrac{\sqrt{3 + x} - \sqrt{1 + x}}{( 3 + x ) - ( 1 + x )}dx$

$⇒ I = \dfrac{1}{2}\left[{∫}_{0}^{1}\sqrt{3 + x}dx - {∫}_{0}^{1} ( \sqrt{1 + x} ) dx\right]$

$⇒ I = \dfrac{1}{2}{\left[\dfrac{2 ( 3 + x) \dfrac{3}{2}}{3} - \dfrac{2 ( 1 + x) \dfrac{3}{2}}{3}\right]}_{0}^{1}$

$⇒ I = \dfrac{1}{2}\left\{\left[\dfrac{2 ( 3 + 1) \dfrac{3}{2}}{3} - \dfrac{2 ( 1 + 1) \dfrac{3}{2}}{3}\right] - \left[\dfrac{2 ( 3) \dfrac{3}{2}}{3} - \dfrac{2 ( 1) \dfrac{3}{2}}{3}\right]\right\}$

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

$⇒ I = \dfrac{1}{2}\left\{6 - \dfrac{4\sqrt{2}}{3} - 2\sqrt{3}\right\}$

$⇒ I = 3 - \sqrt{3} - \dfrac{2}{3}\sqrt{2} = a + b\sqrt{2} + c\sqrt{3}$

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

$⇒ 2a + 3b - 4c = 6 - 2 + 4 = 8$