# Prove the integral

1. May 13, 2008

### inquisitive

$\int du/(u^2-a^2) = (1/2a) ln (u+a)/(u-a) + C$

2. May 13, 2008

### mathwonk

haveyou tried taking the derivative of the RHS?

3. May 14, 2008

### swapnilster

You get the answer by taking u=a*sec y =>y=arcsec u/a(arcsec-sec inverse)
=>du=a sec y tan y dy
Therefore I=integral[(a sec y tan y dy)/a^2 tan^2 y]
=>(sec y dy)/(a tan y)
=>cosec y dy /a
=>ln(cosec y-cot y)/a
then just substitute value of y to get the answer

4. May 14, 2008

### Big-T

You could also do this by partial fraction decomposition.

5. May 14, 2008

### DavidWhitbeck

You need to show that you've attempted the problem.

6. May 15, 2008

### inquisitive

Thanks guys...
The solution swapnilster proposed is correct. I also did his solution before. But after the trigonometric substitution I was stuck at the answer
$\frac{1}{2}ln \frac{(u+a)}{\sqrt{u^2-a^2}}+ C$
This is where I started to need some help from you guys. But after some time I realized to arrive to the form
$\frac{1}{2a}ln \frac{u+a}{u-a}+ C;$
I just need to put the (u+a) term inside the radical and apply some algebra and properties of logarithms. But anyway, thanks again guys.

Last edited: May 15, 2008
7. May 15, 2008

### HallsofIvy

Staff Emeritus
As pointed out by MathWonk, a perfectly valid method of proving that a given function is an anti-derivative is to differentiate it. That, typically, is easier than trying to integrate "from scratch".