Question

Solve:

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

• A. $\frac{4}{3}(\sqrt{2}+1)$
• B. $\frac{3}{4}(\sqrt{2}+1)$
• C. $\frac{3}{4}(\sqrt{2}-1)$
• D. None of these

Answer 1

The correct option is D None of these

We have,

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

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

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

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

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

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

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

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

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

$=\frac{2}{3}[2^{\frac{3}{2}}-1]+\frac{2}{3}\left[1\right]$

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

$=\frac{2}{3}{2}^{\frac{3}{2}}$

Hence, this is the answer.