Question

Integrate:

$\int \sqrt{\frac{1-x}{1+x}}dx$

Answer 1

$I=\int \sqrt{\frac{1-x}{1+x}}dx$

Multiplying by $(1-x)$ and dividing by $(1-x)$

$\Rightarrow I=\int \frac{1-x}{\sqrt{1-x^2}}dx$

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

$\int \frac{x}{\sqrt{1-x^2}}$ is of the form $\int f^{\prime }\left(x\right)dx$

where $f\left(x\right)=\sqrt{1-x^2}$ $\Rightarrow f^{\prime }\left(x\right)=-\frac{x}{\sqrt{1-x^2}}$

$\Rightarrow I={\sin }^{-1}\left(x\right)+\sqrt{1-x^2}+c$