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While going through my text I got stuck in the derivation given in the picture.

enter image description here

($\Omega$ is a constant)

I don't know how to get the second step from the first step, also I don't know why ln is applied in the second step.

Translation (by an editor):

$$\int_{N'}^N dN/N = -\Omega \int_{t'}^t dt$$ $$\ln N-\ln N' = -\Omega (t-t')$$

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  • $\begingroup$ Perhaps you could use the site meta.math.stackexchange.com/questions/5020/… to re-format your question. It isn't readable in this form. $\endgroup$
    – Paul Sundheim
    Aug 21, 2014 at 15:15
  • $\begingroup$ I agree with @Paul. I was about to edit your post, but then realised that it is difficult to understand what you mean. Do you mean the following? $$N^{N^{\prime}_{1/N}}dN=\alpha t^{t^{\prime}_{dt}}$$ This seems odd to me...(If this is what you mean, then the code is $N^{N^{\prime}_{1/N}}dN=\alpha t^{t^{\prime}_{dt}}$) $\endgroup$
    – user1729
    Aug 21, 2014 at 15:28
  • $\begingroup$ I understood that it is difficult to understand the codes I have typed so I have edited the question. $\endgroup$
    – jNerd
    Aug 21, 2014 at 15:33
  • $\begingroup$ Help! is great in the Beatles catalog; not so great in question titles. $\endgroup$
    – user147263
    Aug 21, 2014 at 15:42

1 Answer 1

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The piece you seem to be missing is $\int{\frac{1}{x} dx}=\ln x +C$ Then $\int_M^N {\frac{1}{x} dx}=\ln N-\ln M$ You shouldn't have the same variable inside the integrand as in the limits of the integrand.

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