Question

$\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1+x}-\sqrt{x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\frac{1}{\sqrt{1+x}-\sqrt{x}}\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^1\left(\frac{1}{\sqrt{1+x}-\sqrt{x}}\times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}\right)\mathit{d}x \\ \Rightarrow I={\int }_0^1\frac{\sqrt{1+x}+\sqrt{x}}{1+x-x}\mathit{d}x \\ \Rightarrow I={\int }_0^1\left(\sqrt{1+x}+\sqrt{x}\right)dx \\ \Rightarrow I={\left[\frac{2}{3}{\left(1+x\right)}^{\frac{3}{2}}+\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^1 \\ \Rightarrow I=\frac{2}{3}\times 2\sqrt{2}+\frac{2}{3}-\frac{2}{3} \\ \Rightarrow I=\frac{4\sqrt{2}}{3} \end{gathered}$$