The value of $\int\limits_0^1 \sqrt{\frac{1-x}{1+x}} d x$, is

$\frac{\pi}{2}-1$

We have,

$\int\limits_0^1 \sqrt{\frac{1-x}{1+x}} d x$

$=\int\limits_0^1 \frac{1-x}{\sqrt{1-x^2}} d x=\int\limits_0^1 \frac{1}{\sqrt{1-x^2}} d x-\int\limits_0^1 \frac{x}{\sqrt{1-x^2}} d x$

$=\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1=\pi / 2-1$