Question

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

Answer 1

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

${\int }_0^1\frac{\sqrt{1+x}-\sqrt{x}}{\left(\sqrt{1+x}+\sqrt{x}\right)\left(\sqrt{1+x}-\sqrt{x}\right)}dx={\int }_0^1\frac{\sqrt{1+x}-\sqrt{x}}{\left(1+x-x\right)}dx$

$={\int }_0^1\left(\sqrt{1+x}-\sqrt{x}\right)dx$

$={\int }_0^1{\left(1+x\right)}^{\frac{1}{2}}-{\int }_0^1{x}^{\frac{1}{2}}dx$

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

$=\frac{2}{3}\left[\left(2^{\frac{3}{2}}-1\right)-1\right]=\frac{4(\sqrt{2}-1)}{3}$