# Explanation of the passage from $\int_{N'}^N dN/N$ to $\ln N-\ln N'$

While going through my text I got stuck in the derivation given in the picture. ($\Omega$ 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):

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

• Perhaps you could use the site meta.math.stackexchange.com/questions/5020/… to re-format your question. It isn't readable in this form. Aug 21, 2014 at 15:15
• 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 $N^{N^{\prime}_{1/N}}dN=\alpha t^{t^{\prime}_{dt}}$) Aug 21, 2014 at 15:28
• I understood that it is difficult to understand the codes I have typed so I have edited the question. Aug 21, 2014 at 15:33
• Help! is great in the Beatles catalog; not so great in question titles.
– user147263
Aug 21, 2014 at 15:42

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.