$\sqrt {\cfrac{{1 - \sqrt x }}{{1 + \sqrt x }}}$

: Let $I = \int {\sqrt {\cfrac{{1 - \sqrt x }}{{1 + \sqrt x }}} } dx$

Put $\sqrt x = \cos t$ $\Rightarrow$ $x = {\cos ^2}t$ $\Rightarrow$ $dx = 2\cos t\left( { - \sin \,t} \right)dt$

$\therefore$ $I = \int {\sqrt {\cfrac{{1 - \cos t}}{{1 + \cos t}}} } \left( { - 2\sin \,t\,cos\,t} \right)dt$

$= - 4\int {\sqrt {\cfrac{{2{{\sin }^2}\cfrac{t}{2}}}{{2{{\cos }^2}\cfrac{t}{2}}}} } \sin \cfrac{t}{2}\cos \cfrac{t}{2}\cos t\,dt$

$= - 4\int {{{\sin }^2}\cfrac{t}{2}\cos t\,dt = - 4\int {\cfrac{{1 - \cos t}}{2}} \cos t\,dt}$

$= - 2\int {\left( {\cos t - {{\cos }^2}t} \right)dt} = - 2\int {\left( {\cos t - \cfrac{{1 + \cos 2t}}{2}} \right)} dt$

$= - \int {\left( {2\cos t - 1 - \cos 2t} \right)dt} = - \left[ {2\sin t - t - \cfrac{{\sin 2t}}{2}} \right] + C$

$= - \left[ {2\sin t - t - \sin t\cos t} \right] + C$

$= - \left[ {2\sqrt {1 - x} - {{\cos }^{ - 1}}\sqrt x - \sqrt {1 - x} \sqrt x } \right] + C$

$= - 2\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x + \sqrt x \sqrt {1 - x} + C$