Question

Solve:

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

Answer 1

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

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

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

$1+x=p^2$

$dx=2pdp$

$\Rightarrow {\int }_0^{\sqrt{2}}p\left(2pdp\right)$

(A) $\Rightarrow {\int }_0^{\sqrt{2}}2p^{2}dp$

$\Rightarrow {\frac{2p^3}{3}]}_0^{\sqrt{2}}$

$\Rightarrow \frac{2\times 2\sqrt{2}}{3}$

$A\Rightarrow \frac{4\sqrt{2}}{3}$

(B) ${\int }_0^1\sqrt{x}dx$

$=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}$

$={\frac{2}{3}{x}^{\frac{3}{2}}]}_0^1$

$B=\frac{2}{3}$.