can anyone show me how to integrate √[(1+X)/(1-x)] dx ? im doing revision for my subject and came upon this question where im stuck.
There various ways to go about it.
$\displaystyle \int\sqrt{\frac{1+x}{1-x}}dx$
Make the sub $\displaystyle x=\frac{u^{2}-1}{u^{2}+1}, \;\ dx=\frac{4u}{(u^{2}+1)^{2}}du$
This whittles it down to:
$\displaystyle 4\int\frac{1}{u^{2}+1}du-4\int\frac{1}{(u^{2}+1)^{2}}du$
Now, can you finish it up?.
I would use an algebraic simplification first. Build the fraction using (1+x) to get
$\displaystyle \int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)^2}{1-x^2}}dx=\int\frac{1+x}{\sqrt{1-x^2}}dx=\int \frac{1}{\sqrt{1-x^2}}dx+\int \frac{x}{\sqrt{1-x^2}}dx$
The first is the arcsine and the 2nd is a u sub. I hope this helps.
Note: I didn't use absolute values in the 3rd equality because the domain is $\displaystyle x \in (-1,1)$ so the numerator is always posative there.