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Please confirm simple integration

Okay, we have this:



But I thought log(a) + log(b) = log(ab), not log(a/b)

Or am I being really silly?

Thanks.
Reply 1
Avatar for nuodai
nuodai
17
If you differentiate 1K−N\dfrac{1}{K-N} you get −ln⁡∣K−N∣-\ln |K-N| (since the coefficient of N is -1), hence the result... however it's not quite right, since you should have a constant of integration there (unless the constant on the RHS came from another integral).
nuodai
If you differentiate 1K−N\dfrac{1}{K-N} you get −ln⁡∣K−N∣-\ln |K-N| (since the coefficient of N is -1), hence the result... however it's not quite right, since you should have a constant of integration there (unless the constant on the RHS came from another integral).


I think it looks like a differential equation question, where they have to solve it. It also looks like they have gathered the constants, but didn't change the letter (which is what generally occurs).
It's because:

Edit: People beat me to it =(
mathtex.cgi.gif0.8KB
Reply 4
Avatar for RayW
RayW
OP
Pah, yea, realised it the second I posted it. Hmmm.... I'm not thinking right today. Thanks all.
Reply 5
Avatar for RayW
RayW
OP
nuodai
If you differentiate 1K−N\dfrac{1}{K-N} you get −ln⁡∣K−N∣-\ln |K-N| (since the coefficient of N is -1), hence the result... however it's not quite right, since you should have a constant of integration there (unless the constant on the RHS came from another integral).

Yea, the rhs came from another integral.
Reply 6
Avatar for Slumpy
Slumpy
16
nuodai
If you differentiate 1K−N\dfrac{1}{K-N} you get −ln⁡∣K−N∣-\ln |K-N| (since the coefficient of N is -1), hence the result... however it's not quite right, since you should have a constant of integration there (unless the constant on the RHS came from another integral).


Careful:p:
Slumpy
Careful:p:


We all knew what he meant :p:

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