Originally Posted by **topsquark**
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.