I have come across the following operation in two electrodynamics textbooks, which I find problematic: When evaluating an integral over a Laplacian in a spherically symmetric function, the radial term $\frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2 \frac{\partial \Phi}{\partial r} \right)$ is evaluated simply as $\frac{\partial^2\Phi}{\partial r^2}$, i.e. $$ \int_\mathcal{V} \mathrm{d}V\text{ }\frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2 \frac{\partial \Phi}{\partial r} \right) = \int_\mathcal{V}\mathrm{d}V\text{ }\frac{\partial^2\Phi}{\partial r^2}. $$
How is this true?
The examples I saw were from J. B. Marion's Classical Electromagnetic Radiation and Andrew Zangwill's Modern Electrodynamics. I'll give an example from the former for context:
To prove that the retarded potential $$ \Phi(\mathbf{r}, t) = \int_\mathcal{V} \mathrm{d}V'\text{ }\frac{\rho (\mathbf{r}', t-|\mathbf{r}-\mathbf{r}'|/c)}{|\mathbf{r}-\mathbf{r}'|} $$ satisfies $$ \operatorname{\Box} \Phi (\mathbf{r},t) = -\frac{\rho (\mathbf{r},t)}{\epsilon_0}, $$ we separate $\mathcal{V}$ into a small region around $\mathbf{r}$ that tends to $0$, $\mathcal{V}_1$, and $\mathcal{V}_2 = \mathcal{V}-\mathcal{V}_1$. We can then solve for $\Phi_1$ and $\Phi_2$ over the two regions such that $\Phi = \Phi_1+\Phi_2$ with suitable approximations.
The $ \nabla^2 \Phi_1$ term gives $- \rho (\mathbf{r},t) / \epsilon_0 $, while the $\nabla^2\Phi_2$ term is such that $$ \nabla^2\Phi_2 (\mathbf{r},t) = \frac{1}{4\pi\epsilon_0}\int_{\mathcal{V}_2}\mathrm{d}V'\text{ }\nabla^2\left(\frac{\rho (\mathbf{r}',t-|\mathbf{r}-\mathbf{r}'|/c)}{|\mathbf{r}-\mathbf{r}'|}\right). $$
Then the author makes use of the operation I mentioned above to turn this into $$ \nabla^2\Phi_2 = \frac{1}{4\pi\epsilon_0}\int_{\mathcal{V}_2}\mathrm{d}V'\text{ }\frac{1}{R}\frac{\partial^2\rho (\mathbf{r}',t-R/c)}{\partial R^2}, $$ where $R = |\mathbf{r}-\mathbf{r}'|$. By virtue of the $t-R/c$ dependence, we know that $\rho(\mathbf{r},t)$ satisfies the wave equation with speed $c$, so the spatial derivative can be replaced by a temporal derivative, thus completing the proof by noticing that $$ \nabla^2\Phi_2 = \frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2} $$ in the limit as $\mathcal{V}_1 \to 0$.
