Where Do the Factors c^2 and ab Originate in the Coaxial Cable Integral?

[Lavace]

http://img154.imageshack.us/img154/165/integrallll.jpg

After working it out myself, I have the logarithim part equal to:

[ln b - lna] + [ln c - ln b] = [ln b/a + ln c/b] = ln c/a (Not sure about this statement!)

I don't understand where the c^2 and ab came from with this integral?

Any help?

Thanks!

 

[gabbagabbahey]

After working it out myself, I have the logarithim part equal to:

[ln b - lna] + [ln c - ln b] = [ln b/a + ln c/b] = ln c/a (Not sure about this statement!)

I don't understand where the c^2 and ab came from with this integral?

Any help?

Thanks!

Take a careful look at the factors in front of each logarithm (they are different by a factor of 2) and remember that p\ln(x)=\ln(x^p)