quantum mechanics - Expectation value of $p^2 (1/r) + (1/r) p^2$

I'm trying to derive $\langle kl | \lbrace 1/r , p^2 \rbrace | nl \rangle$, where the states satisfy the equation of motion (I omit factors of $1/2m$ etc.):

$$(p^2 + V)| n,l \rangle = E_n | n,l \rangle$$

and $\lbrace A,B \rbrace = AB + BA$ is the anticommutator. I have two solutions at hand that differ. In general I can write:

$$\langle k,l | \lbrace 1/r , p^2 \rbrace | n,l \rangle = \langle k,l | 1/r~p^2 + p^2~1/r | n,l \rangle$$

and I can use that $p^2 \propto -\nabla^2$ and $\nabla^2 1/r \propto -\delta^{(3)}(r)$.

Now I want to let $p^2$ act first on the respective states and let $1/r$ act later on, such that I can use the e.o.m. Thus I have:

$$\langle k,l | 1/r~p^2 + p^2~1/r | n,l \rangle = \langle k,l | 1/r~(E_n - V) + (E_k - V)~1/r | n,l \rangle$$

In this solution I let the first summand act on the ket and I let the second summand act on the bra. However, when I do it the other way around, using the product rule $\nabla^2 (1/r~\psi) = (\nabla^2 1/r) \psi + 1/r (\nabla^2 \psi)$ and use the e.o.m for the second term I obtain: \begin{eqnarray} \langle k,l | 1/r~p^2 + p^2~1/r | n,l \rangle &=& \langle k,l | \delta^{(3)}(r) + (E_k - V) + (E_n - V) + \delta^{(3)}(r) | n,l \rangle\\ &=& \langle k,l | 1/r~(E_n - V) + (E_k - V)~1/r | nl \rangle + 2 \langle kl | \delta^{(3)}(r) | n,l \rangle \end{eqnarray}

These two solutions are clearly different if and only if both of the wave functions do not vanish at the origin. This is to say that these two solutions give different results for s-waves.

Am I missing something essential here? Any input would be appreciated!