Given any distribution $(\mathcal{D}')$ its distributional derivative (which is again a distribution) always exists. In the other hand, do we always have antiderivative?
On other words, given $v\in \mathcal{D}'(\Omega)$ does the PDE $u_x=v$ have a distributional solution?
If not, under what assumptions on $v$ does the pde have a solution.
I will give the solution for $\Omega\subseteq \mathbb{R}$ non-empty open connected (interval). I suspect that the general multidimensional case is more difficult.
If $u$ as in your statement exists, then $$<u, \varphi_x>=-<v, \varphi>$$ for every $\varphi \in \mathcal{D}(\Omega)$. Let us "shift the problem" on the test functions, i.e. determine when, given a certain $\varphi$, the equation $$\psi_x=\varphi$$ has a solution $\psi \in \mathcal{D}(\Omega)$. It is obvious from the fundamental theorem of calculus that $$\int_\Omega \varphi=0$$ is a necessary condition. It is also sufficient since if it holds then $$\psi(x)=\int_{\inf \Omega}^x \varphi(t) dt$$ is a solution (being supported in the biggest compact interval containing supp $\varphi$). Therefore, if we fix $\varphi_0 \in \mathcal{D}(\Omega)$ s.t. $\int_\Omega \varphi_0=1$ then for every $\varphi \in \mathcal{D}(\Omega)$ $$<u, \varphi-(\int_\Omega \varphi)\varphi_0>=-<v,\psi>$$ and so $$<u, \varphi>=\int_\Omega \varphi <u, \varphi_0>-<v, \psi>$$ where $$\psi(x)=\int_{\inf \Omega}^x [\varphi(t)-(\int_\Omega \varphi)\varphi_0(t)]dt$$ Fixed a value for $<u, \varphi_0>$, this gives the expression of a well-defined linear and continuous functional $u \in \mathcal{D}'$ which satisfies the equation. So a solution always exists (and is unique up to a constant).