Question

Solve:

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

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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