Question

Evaluate the definite integrals.
${\int }_0^1\frac{1}{\sqrt{1+x}-\sqrt{x}}dx.$

Answer 1

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