In my EMF lecture, the professor first formulated the equation below for an infinitesimal part of a force and then integrated it. I can't explain this step with what I know from my math lectures. When integrating an equation, I would expect a '$\int_a^b dx$' (or some vectorial form of that) to be added to both sides but he just added the integral sign (no additional differential!). Furthermore, I would expect a change of dimension as a result of the integration (like e.g. acceleration integrated results in velocity) - accordingly, an integration of a force would not result in a force again. How can I explain what he did? I suspect that it is probably shorthand for some other integration.
$$d\vec{F}_{PP'}=\frac{1}{4\pi\varepsilon_0}\varrho(\vec{r}')\varrho(\vec{r})\frac{\vec{r}'-\vec{r}}{\lvert\vec{r}'-\vec{r}\rvert^3}dVdV'=-d\vec{F}_{P'P} \\ \implies \vec{F}=\frac{1}{4\pi\varepsilon_0}\int_V\int_V\varrho(\vec{r})\varrho(\vec{r}')\frac{\vec{r}-\vec{r}'}{\lvert\vec{r}-\vec{r}'\rvert^3}dV'dV=\vec{0}$$
Edit To simplify the question, essentially what I am asking is:
I know that $$ f(x) = g(x) \implies \int_a^b f(x) \,dx = \int_a^b g(x) \,dx. $$
But how do I explain that is valid: $$ df = g(x)\,dh \implies f = \int_a^b g(x) \,dh $$