I have been trying to get my head around this for some time now... I solve the same integral in two ways but get two different solutions. Since there can't (surely) be any sort of ambiguity when integrating, the answers have to either be identical (ruled out) or I am doing something wrong in one of the solutions. My theory is that the missing ln(1/k) is embodied, somehow, in the constant. Could this be it?

Method 1: $$ \int \frac{1/k}{1-y/k}dy=\frac{1}{k}\int \frac{1}{1-y/k}dy\rightarrow v = 1-y/k\rightarrow =-ln(1-y/k) + c $$

Method 2: $$ \int \frac{1/k}{1-y/k}dy = \int \frac{1}{k-y}dy=-ln(k-y) + c $$

Many thanks in advance V.Vocor

  • 4
    $\begingroup$ You're right. If you take the difference of the two results, you get $\ln k$ plus a constant, which results in a constant. $\endgroup$ Sep 24 '14 at 15:31

$$\ln(k-y)+c=\ln((1-y/k)k)+c=\ln (1-y/k)+\underbrace{\ln k+c}_{c'}$$ Use $\ln ab=\ln a+\ln b$

  • $\begingroup$ Words? Help the OP here. $\endgroup$ Sep 24 '14 at 16:58
  • $\begingroup$ @Alizter seetheedit $\endgroup$
    – RE60K
    Sep 24 '14 at 17:03

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