The integral $\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x$ is equal to (where C is the constant of integration)

$-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Let $x=\cos ^2 \theta$. Then, $d x=-2 \sin \theta \cos \theta d \theta$.

∴   $I=\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x$

$\Rightarrow I =\int \frac{1}{(1+\cos \theta) \cos \theta \sin \theta} \times(-2 \sin \theta \cos \theta) d \theta$

$\Rightarrow I=-\int \sec ^2 \frac{\theta}{2} d \theta=-2 \tan \frac{\theta}{2}+C$

$\Rightarrow I=-2 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}+C=-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$