Question

Show that $\int \sqrt{\left(\frac{1+x}{1-x}\right)}dx={\sin }^{-1}x-\sqrt{1-x^2}$.

Answer 1

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

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

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

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

Put $1-x^2=t$ in the second integral

$\Rightarrow -2xdx=dt$

$I={\sin }^{-1}x-\frac{1}{2}\int \frac{1}{\sqrt{t}}dt$

$={\sin }^{-1}x-\frac{1}{2}\frac{\sqrt{t}}{\frac{1}{2}}+C$

$I={\sin }^{-1}x-\sqrt{1-x^2}+C$