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Old Dec 21st 2013, 07:46 PM   #21
Pmb
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Join Date: Apr 2009
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Originally Posted by topsquark View Post


Let v = sqrt(g/k) * u. Then dv = sqrt(g/k) du. The integral becomes:


So


Which you can finish from here. (The solution for v can be put into the form of an inverse hyperbolic tangent for a "cleaner" solution, but I think the log format is easier to grasp.)

-Dan
I thought it would be helpful to show how that integral was evaluated. The method is called expansion by patial fractions. We write 1/(1 - u^2) as

1/(1 - u^2) = (1/2)[ 1/(1 - u) - 1/(1 + u) ]

When this is substituted its easy to see how to evaluate it in terms of natural logarithms.
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