Prove the integral

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

Homework Helper
haveyou tried taking the derivative of the RHS?

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

Big-T
You could also do this by partial fraction decomposition.

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

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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.

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