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I'm reading a book in electrical engineering and in one example they are skipping steps and I'm not sure how they integrate the equation: $$ \frac{dv}{dt} + \frac{v}{RC} = 0 $$ which they rewrite as: $$ \frac{dv}{v} = -\frac{dt}{RC}$$

Integrating the function gives: $$ \ln(v) = -\frac{t}{RC} + \ln(A) $$ Where $A$ is the constant. $\frac{1}{v}$ gives $\ln(v)$, but I don't get the right hand side. Why not just '$A$'?

Thanks.

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    $A$ and $\ln(A)$ are the same, since A can be anything (including a complex number). They are using $\ln(A)$ so the next step of raising everything to an exponential is more clear. I personally wouldn't do it like that. I would instead note that $e^A$ is the same as $A$ in the next step. – 2017-01-07
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    a constant is a constant – 2017-01-07
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    Okay, thanks. It makes sense, it's been a while since I did calculus. – 2017-01-07

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The author is thinking ahead in the problem. If we write $\ln(v) = -\frac{t}{RC} + B$, we can then exponentiate both sides of this equation: $$e^{\ln(v)} = e^{-\frac{t}{RC}} e^B$$ so if $B=\ln(A)$ then $$v = A e^{-\frac{t}{RC}}$$ We can always find some $A$ for which $B=\ln(A)$ because the range of $\ln$ is all real numbers. The only reason they did this was to simplify the final result. It isn't a spooky integration trick, don't worry.

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Well, it helps to introduce the constant of integration in this way because when you look at the LHS it is ugly. But seriously by introducing it this way you can learn a lot more about the system for e.g. you can rewrite it as $ln(v)-ln(A)=-t/RC$ which can be re-written as $v/A=e^{-t/RC}$ and hence you can see $v/a$ decays over time or whatever it may be. Make sense?