Question

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

• A. $\frac{2\sqrt{2}}{3}$
• B. $\frac{4\sqrt{2}}{3}$
• C. $\frac{8\sqrt{2}}{3}$
• D. None of these

Answer 1

The correct option is D None of these

$I={\int }_0^1\frac{dx}{\sqrt{1+x-\sqrt{x}}}$ let $\sqrt{x}=u\Rightarrow du=\frac{1}{2\sqrt{x}}dx$

when $x=0,4=0$ & $x==1.u=1$

$={\int }_0^1\frac{2udu}{\sqrt{1+u^2-u}}$

$={\int }_0^1\frac{(2u-1)+1}{\sqrt{u^2-u+1}}du={\int }_0^1\frac{(2u-1)}{\sqrt{u^2-u+1}}du+{\int }_0^1\frac{du}{\sqrt{u^2-u+1}}$

$={\left[2\sqrt{u^2-u+1}\right]}_0^1+{\int }_0^1\frac{du}{\sqrt{{(u-\frac{1}{2})}^2+\frac{3}{4}}}[\because \frac{dx}{\sqrt{x^2+a^2}}=\log |x+\sqrt{x^2+a^2}|+c]$

$=2(1-1)+{\left[\log |u-\frac{1}{2}+\sqrt{{(u-\frac{1}{2})}^2+\frac{3}{4}}|\right]}_0^1$

$=\log |1-\frac{1}{2}+1|-\log |1-\frac{1}{2}|$

$=\log 3$

$\therefore$ option $D$ is correct.