<s>[INST] 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? 
 [/INST] 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$ 
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