How would I write this without separation of variables? I'm reading an explanation of how to solve first-order differential equations.  Part of the way through, I have this:
If $\frac{dR}{dx}=RP$,
$$\begin{align*}
\frac{dR}{R} &= Pdx, \\
\int \frac{dR}{R} &= \int Pdx, \\
\ln R &=\int Pdx +c.
\end{align*}
$$
Now, this makes me uneasy: I don't like to separate variables.  I do it when I integrate with a substitution, but I'm perfectly aware when I'm doing it that I've slipped out of math for a second to manipulate my symbols for convenience, and I could do it more formally if I had to.
That's what I want to do here, but I can't figure it out.  Given the first line, how do I solve for R without separation of the variables?
I should add that P and R are both functions of x.
 A: I had this question before too, and I looked through Art of Problem Solving's Calculus textbook at one point to see if they had a more formal way of doing it. I believe this is more or less what I came across:
We can divide both sides by $R$, which gives
$\dfrac{1}{R} \dfrac{dR}{dx} = P$.
This can be integrated with respect to $x$ which gives
$\displaystyle \int \dfrac{1}{R} \dfrac{dR}{dx} dx = \int P dx$.
Then by the Chain Rule, the LHS equals $\displaystyle \int \dfrac{1}{R} dR$, and the remainder of the integration can be carried out. Wikipedia (http://en.wikipedia.org/wiki/Separation_of_variables) mentions that this last step is due to the "substitution rule for integrals." 
A: You could notice that you have the derivative of a function equaling the function itself scaled by a constant.
Then, you know that only a certain class of functions has that property, namely $e^x$. You could then assume that $R = e^{kx}+C$, and plug it into the equation and solve for $k$.
Somehow, this doesn't seem more satisfying, however.
A: Well, you haven't slipped out of math at all. Actually, this kind of Newtonian manipulation of ${\rm d}x$ works very often and can be made rigorous if necessary.
Here, observe that if the differential equation $${R'(x) \over R(x)} = P(x)$$ holds on some interval $[a, b]$ then it is also certainly the case that $$\int_a^y {R'(x) \over R(x)} {\rm d} x = \int_a^y P(x)$$ for any $y \in [a,b]$.
Now use the substitution $u = \log(R(x))$, ${\rm d}u = {R' {\rm d}x \over R}$
to get $$\log(R(y)) + C = \int_{\log(R(a))}^{\log(R(y))} {\rm d} u = \int_a^y P(x)$$ again for all $y \in [a,b]$ and so this can be also translated into the language of indefinite integral by the fundamental theorem of calculus.
In other words, that juggling of infinitesimals is actually substitution under the integral sign (and in particular valid as long as hypothesis of substitution theorems are satisfied).
