Posted by Anthony on Thursday, December 6, 2007 at 1:05am.
The indefinite integral of
dv/(z^2-v^2),
with z being a constant, is
[1/(2z)]log[(z+v)/(z-v)]
Evaluate that at v=v' and subtract the value for v=0, to get the definite integral.
The method of partial fractions can used to get the integral. It involves rewriting 1/[(z^2-v^2} as
[1/(2z)][1/(z+v) - 1/(z-v)]
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