Question

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

Answer 1

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

Rationalising the denominator

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

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

${\left(\frac{2\sqrt{(1+x{)}^3}}{3}\right)}_0^1+{\left(\frac{2x^{\frac{3}{2}}}{3}\right)}_0^1$

$\frac{4\sqrt{2}}{3}-\frac{2}{3}+\frac{2}{3}=\frac{4\sqrt{2}}{3}$