Question

${\int }_0^1\frac{1}{\sqrt{1+x}+\sqrt{x}}dx=\frac{4}{3}[\sqrt{k}-1]$ Find the value of k

Answer 1

$\int \frac{1}{\sqrt{1+x}+\sqrt{x}}dx=\int =\frac{\sqrt{1+x}-\sqrt{x}}{(\sqrt{1+x}+\sqrt{x})(\sqrt{1+x}-\sqrt{x})}dx={\int }_0^1(\sqrt{1+x}-\sqrt{x})dx$
$={\left[\frac{2}{3}\right(1+x{)}^{\frac{3}{2}}-\frac{2}{3}{x}^{\frac{3}{2}}]}_0^1$
$=\left[\frac{2}{3}\right(1+1{)}^{3/2}-\frac{2}{3}\left(1{)}^{3/2}\right]-=\left[\frac{2}{3}\right(1+0{)}^{3/2}-\frac{2}{3}\left(0{)}^{3/2}\right]$
$\frac{2}{3}\left[2^{\frac{3}{2}-1}\right]-\frac{2}{3}[1+0]=\frac{2}{3}[2\sqrt{2}-2]=\frac{4}{3}[\sqrt{2}-1]$