Question

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

• A. $\frac{4}{3}(\sqrt{2}-1)$
• B. $\frac{3}{4}(\sqrt{2}-1)$
• C. $\frac{4}{3}(1-\sqrt{2})$
• D. $\frac{3}{4}(1-\sqrt{2})$

Answer 1

The correct option is A $\frac{4}{3}(\sqrt{2}-1)$

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

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

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

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

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

$=\frac{2}{3}[2^{3/2}-1]-\frac{2}{3}[1-0]$

$=\frac{2}{3}(2\sqrt{2}-1)-\frac{2}{3}$

$=\frac{2}{3}[2\sqrt{2}-1-1]$

$=\frac{2}{3}(2\sqrt{2}-2)=\frac{4}{3}(\sqrt{2}-1)$.