$\int \sqrt{\frac{1+x}{1-x}} d x=$ (where $C$ is a constant of integration.)

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int \sqrt{\frac{1+x}{1-x}} \mathrm{~d} x=\int \sqrt{\frac{(1+x)^2}{1-x^2}} \mathrm{~d} x=\int\left\{\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}ight\} \mathrm{d} x \ & =\sin ^{-1} x-\frac{1}{2} 	imes 2 \sqrt{1-x^2} \ & =\sin ^{-1} x-\sqrt{1-x^2}+C\end{aligned}$

$\int \sqrt{\frac{1+x}{1-x}} d x=$ (where $C$ is a constant of integration.)

Solution