I have asked questions before, and have read many questions and answers from others on stackexchange, about how to treat "$dx$" in a differential equation.
people often have the intuition of treating $dx$ in the term "$\frac{df}{dx}$" as if it were a separate subterm of that, even though $\frac{df}{dx}$, or at least $\frac{d}{dx}$ is an operator, and shouldn't be seen as a fraction of two seperable elements.
Nevertheless, when solving the simple differential equation $\frac{dx}{dt}=x$, people often proceed as follows:
$$(a):\qquad \frac{dx}{dt}=x$$ Step 1: multiply by $dt$ and subtract by $x$: $$(b):\qquad \frac{1}{x}dx=dt $$ Step 2: integrate both sides: $$(c): \qquad \int\frac{1}{x}dx=\int dt$$ Step 3: Solve the integral: $$(d): \qquad ln(x)=t+C\implies x=e^{t+C}=x_0e^t$$
So my questions are
we know that the justification of Step 1 as "multiplication" is incorrect, since $dx$ and $dt$ are not seperable elements, so what is the justification for going from equation $(a)$ to $(c)$?
When $a=b$, we can conclude that $\int adx=\int bdx$, since this takes the integral of both sides with respect to the same variable, but What is the justification of inserting the integral sign in equation $(c)$, without also adding the differential $dx$ to both sides?
The proof hops over (b) and goes straight from (a) to (c).
Generally speaking we have (if $f(x)\not=0$): $$\dfrac{dx}{dt}=f(x)g(t)\Leftrightarrow \int\dfrac{1}{f(x)}dx = \int g(t)\,dt$$ since differentiating the RHS with respect to $t$ gives $$\dfrac{1}{f(t)}\dfrac{dx}{dt}=g(t)$$ according to the chain rule.