Integral exercise asked in math group

Exercise:

Integrate, $\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,\frac{dx}{x}}$

Solution: Let ${{u}^{2}}=x\Leftrightarrow 2udu=dx$

So $\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,\frac{dx}{x}}=\int{\sqrt{\frac{1-u}{1+u}}\times \frac{2udu}{{{u}^{2}}}}=\int{\sqrt{\frac{1-u}{1+u}}}\times \frac{2du}{u}$

But $\sqrt{\frac{1-u}{1+u}}=\frac{\sqrt{1-u}}{\sqrt{1+u}}\times \frac{\sqrt{1-u}}{\sqrt{1-u}}=\frac{1-u}{\sqrt{1-{{u}^{2}}}}$

So $\int{\sqrt{\frac{1-u}{1+u}}\times \frac{2du}{u}}=2\int{\frac{1-u}{u\sqrt{1-{{u}^{2}}}}du}=2\int{\frac{du}{u\sqrt{1-{{u}^{2}}}}-2\int{\frac{u}{u\sqrt{1-{{u}^{2}}}}du}}$

$=2\int{\frac{du}{u\sqrt{1-{{u}^{2}}}}}-2\int{\frac{du}{\sqrt{1-{{u}^{2}}}}}$  take  $u=\sin \theta \Leftrightarrow du=\cos \theta \,d\theta $

$=2\int{\frac{\cos \theta d\theta }{\sin \theta \sqrt{1-{{\sin }^{2}}\theta }}-2\int{\frac{\cos \theta d\theta }{\sqrt{1-{{\sin }^{2}}\theta }}}}=2\int{\frac{1}{\sin \theta }d\theta -2\int{d\theta }}=2\int{\csc \theta \,d\theta }-2\theta +c$

But $\int{\csc \theta \,d\theta }=\int{\frac{\csc \theta \left( \csc \theta -\cot \theta  \right)}{\csc \theta -\cot \theta }d\theta }=\int{\frac{d\left( \csc \theta -\cot \theta  \right)}{\csc \theta -\cot \theta }d\theta }=\ln \left| \csc \theta -\cot \theta  \right|+c$

But $\csc \theta =\frac{1}{\sin \theta }=\frac{1}{u}\,\,\,\And \,\,\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{u}^{2}}}$

Thus $\int{\frac{2\left( 1-u \right)}{u\sqrt{1-{{u}^{2}}}}du}=2\ln \left| \frac{1}{u}-\frac{\sqrt{1-{{u}^{2}}}}{u} \right|-2\arcsin u+c$

Therefore, $\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\frac{dx}{x}}=2\ln \left| \frac{1-\sqrt{1-x}}{\sqrt{x}} \right|-2\arcsin \sqrt{x}+c$