1708\hspace{-20}$Given $\bf{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx =\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\cdot \sqrt{\frac{1-\sqrt{x}}{1-\sqrt{x}}}dx= \int \frac{1-\sqrt{x}}{\sqrt{1-x}}}dx.............$\\\\\\ Let $\bf{x=\cos^2 \phi\;,}$ Then $\bf{dx = 2\cos \phi \cdot -\sin \phi d\phi}$\\\\\\ So Integral is $\bf{=-\int \frac{\left(1-\cos \phi \right)}{\sqrt{1-\cos^2 \phi}}\cdot 2\sin \phi \cdot \cos \phi d\phi}$\\\\\\ So Integral is $\bf{ = \int \left\{2\cos^2 \phi -2\cos \phi \right\}d\phi}$\\\\\\ So Integral is $\bf{ = \int \left(1+\cos 2\phi -2\cos \phi\right)d\phi}$\\\\\\ So Integral is $\bf{=\phi+\frac{\sin 2\phi}{2}-2\sin \phi+\mathbb{C}}$\\\\\\ So $\bf{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx = \cos^{-1}\left(\sqrt{x}\right\;)+\sqrt{x-x^2}-2\sqrt{1-x}+\mathbb{C}}$