How Can Substitution Simplify the Integral of \(\sqrt{\frac{1+x}{1-x}}\) dx?

[suspenc3]

Hi, I am kinda stuck on the following integral:


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

any hints?

 

[TD]

If you let

<br /> y^2 = \frac{{1 + x}}{{1 - x}} \Leftrightarrow x = \frac{{y^2 - 1}}{{y^2 + 1}}<br />

The integral will become fraction of rationals, losing the square root.

 

[suspenc3]

so are you saying to substitute that for x?

 

[suspenc3]

or by making this substitution, the square root will be taken away

 

[neutrino]

Or you could do the trig substitution x = \sin\theta

 

[TD]

so are you saying to substitute that for x?

Yes, use that substitution to lose the square root.

I already solved for x as well, which allows you to easily find dx in terms of dy by differentiating both sides.