Question

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

Answer 1

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