1
$\begingroup$

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

$\endgroup$
  • 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. $\endgroup$ – Milind Sep 24 '14 at 15:31
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.