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

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Lavace
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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!
 
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Lavace said:
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 [itex]p\ln(x)=\ln(x^p)[/itex]