# Thread: first order differential equation

1. ## first order differential equation

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.

2. 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?.

3. Originally Posted by galactus

Make the sub $\displaystyle x=\frac{u^{2}-1}{u^{2}+1}, \;\ dx=\frac{4u}{(u^{2}+1)^{2}}du$
im can solve it once you gave me that... but how do you know what to sub it with?

4. Originally Posted by yen yen
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.

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.