Question

Evaluate ${\int }_0^1\frac{1}{\sqrt{1+x}-\sqrt{x}}dx$

• A. $\frac{4\sqrt{2}}{3}$
• B. $2\sqrt{2}-2$
• C. $\frac{4}{3}[\sqrt{2}-1]$
• D. $\frac{2}{3}[2\sqrt{2}-1]$

Answer 1

The correct option is A $\frac{4\sqrt{2}}{3}$

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