How do I integrate $\sqrt{\frac{1+x}{1-x}}$? How do I integrate $\sqrt{\frac{1+x}{1-x}}$ using standard calculus techniques?
I tried trig substitution but it doesn't seems to work. is that some kind of u-substitution? 
 A: To start, rationalize the numerator:
$$\begin{align}
\int\sqrt{{1+x\over 1-x}}\,dx & =\int\frac{1+x}{\sqrt{1-x^2}}\,dx\\
& =\int\frac{dx}{\sqrt{1-x^2}}+\int\frac x{\sqrt{1-x^2}}\,dx.
\end{align}$$
On the last line, you can use trig substitution on the first integral and $u$-substitution on the second.
A: To find $ \int \sqrt{\frac{1+x}{1-x}}dx $
You begin by multiplying $\frac{\sqrt{1+x}}{\sqrt{1+x}}$ to the integral.
then you get 
$ \int \sqrt{\frac{(1+x)^2}{1-x^2}}dx $
now use trig substitution of 
$x=sin \theta$
$dx = cos \theta d\theta$
we get $ \int \sqrt{\frac{(1+sin \theta)2}{1-sin^2 \theta}}cos \theta \, d\theta = \int 1+sin \theta \, d\theta = \theta - cos\theta + C = arcsin x - \sqrt{1-x^2} + C$ 
A: $$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=|\cot\frac{\theta}{2}|$$
let $$x=\cos \theta$$
$$dx=-\sin \theta d\theta$$
$$\int \sqrt{\frac{1+ x}{1- x}}dx= \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}(-\sin \theta d\theta)=\int |\cot\frac{\theta}{2}|(-\sin \theta d\theta)$$
$$\int |\frac{1+\cos \theta}{\sin \theta}|(-\sin \theta d\theta)=-\int(|1+\cos \theta|)d\theta$$
