Poisson's equation and Green's functions

Trying to verify the general integral solution to Poisson's equation I reach the following contradiction, what am I missing/doing wrong?

A solution $u(\mathbf{x})$ to Poisson's equation satisfies:

$\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}$

we can find such a solution in a given domain by evaluating:

$u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}$

or equivalently (if we are only interested in the electric field):

$\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}$

We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density $\rho(\mathbf{x})$ .

However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of $-\frac{\rho}{\epsilon_0}$ . What am I missing??

To evaluate the Laplacian/divergence of the above I make use of the fact that $\mathbf{x'}$ is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying $\nabla^2$ to the integral on the RHS and also means the components of $\mathbf{x'}$ all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.

EDIT: OK it's since been pointed out to me (at almost the same time as I found it in a book myself --- isn't it always the way) that my error lies in thinking the Laplacian of $\frac{1}{|x-x'|}$ is zero when it fact it is equal to $-4\pi\delta(x-x')$ , which then allows you to get the correct result in the above. However I'm still wondering if anyone knows how to show this is the case directly (i.e. by performing the differentiation), as the references I've found instead start by searching for a function $G$ that satisfies $\nabla^2 G = \delta(x-x')$ and then deducing that it must be the above without direct differentiation.