$\begin{aligned} & \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=\int \sqrt{\frac{(1-\sqrt{x})^2}{1-x}} d x \\ & \int \frac{1-\sqrt{x}}{\sqrt{1-x}} d x=\int \frac{d x}{\sqrt{1-x}}-\int \frac{\sqrt{x}}{\sqrt{1-x}} d x \\ & -2 \sqrt{1-x}-\int \frac{\sqrt{\cos ^2 \theta}}{\sqrt{1-\cos ^2 \theta}}(-2 \cos \theta \cdot \sin \theta d \theta)\left[\operatorname{let} x=\cos ^2 \theta\right] \\ & =-2 \sqrt{1-x}+\int 2 \cos ^2 \theta d \theta \\ & \end{aligned}$
$\begin{aligned} & =-2 \sqrt{1-x}+\int(1+\cos 2 \theta) d \theta \\ & =-2 \sqrt{1-x}+\theta+\frac{\sin 2 \theta}{2}+C \\ & =-2 \sqrt{1-x}+\theta+\sin \theta \cdot \cos \theta+C \\ & =-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}+C \\ & =-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C\end{aligned}$