What is the inverse of the *divergence* operator? The inverse of derivation is integral.
But what is the inverse of the divergence operator ? Doest it exist ?
 A: The answer by Keith is close, except note that the divergence operator is not invertible, just like the derivative. It's "inverse" would also have some degrees of freedom.
In particular, when inverting the derivative $F'=f$, we have $F(y)=\int_{x=0}^{y} f(x)  dx +C$.
If instead, we want to solve $\nabla \cdot \boldsymbol{F}=f$, we have $\boldsymbol{F}(r)=\boldsymbol{F}_0(f)(r)+\boldsymbol{C}(r)$ where $\boldsymbol{F}_0$ is an operator that takes $f$ and applies the Coulomb's integral. 
$\boldsymbol{C}$ can be any function with a constant divergence of $0$. Written in unconstrained form $\boldsymbol{C}=\nabla \times \boldsymbol{J}$ for any vector field $\boldsymbol{J}(r)$.    
A: For example, the charge density is proportional to the divergence of the electric field (Gauss's Law):
$$\frac{\rho}{\varepsilon_0} = \nabla \cdot \mathbf{E}$$
The inverse is Coulomb's Law (for continuous charge distributions): 
$$\boldsymbol{E}(\boldsymbol{r}) = {1\over 4\pi\varepsilon_0}\int \rho(\boldsymbol{r'}) {\boldsymbol{r} - \boldsymbol{r'} \over |\boldsymbol{r} - \boldsymbol{r'}|^3} d^3r'$$
Wikipedia
