Momentum commutation for boson field

Given a boson field described by $$\psi(\vec{x})$$, conserved momentum from the Lagrangian (which isn't relevant here) is $$\vec{P} = \frac{\hbar}{2i} \int d^3 x \left( \psi^\dagger \nabla \psi - \nabla \psi^\dagger \psi \right)$$. Question claims the commutator of $$\vec{P}$$ and $$\psi(\vec{x})$$ is $$\frac{\hbar}{i} \nabla \psi (\vec{x})$$ but my math doesn't lead me there. Expanding the commutator results in:

$$\left[ \vec{P}, \psi (\vec{x}) \right] = \frac{\hbar}{2i} \int d^3 x \left[ \psi^\dagger (\vec{x}') \nabla \psi (\vec{x}') - \nabla \psi^\dagger (\vec{x}') \psi (\vec{x}'), \psi(\vec{x} ) \right]$$ $$= \frac{\hbar}{2i} \int d^3 x' \left[ \psi^\dagger (\vec{x}'), \psi (\vec{x} ) \right] \nabla \psi (\vec{x}') + \frac{\hbar}{2i} \int d^3 x' \psi^\dagger (\vec{x}') \left[ \nabla \psi (\vec{x}') , \psi (\vec{x} ) \right] - \frac{\hbar}{2i} \int d^3 x' \left[\nabla \psi^\dagger (\vec{x} '), \psi (\vec{x}) \right] \psi (\vec{x}') - \frac{\hbar}{2i} \int d^3 x' \nabla \psi^\dagger (\vec{x} ') \left[\psi (\vec{x}'), \psi (\vec{x})\right]$$

The first term turns into a delta function that becomes $$-\frac{\hbar}{2i} \nabla \psi (\vec{x})$$ and the last term is zero, but I'm not sure how to evaluate the middle two terms.

For the middle terms you use that the nabla differential acts on fields indexed by $$\vec x^\prime$$. It sees the field operator $$\psi(\vec x)$$ as a constant due to its independence of $$\vec x^\prime$$. Thus, we can pull the differential operator out of the commutator which yields for example $$\int d^3\vec x^\prime \psi(\vec x^\prime) \nabla_{\vec x^\prime} [\psi(\vec x^\prime), \psi(\vec x)]$$ for the second term. The subscribt $$\vec x^\prime$$ of $$\nabla_{\vec x^\prime}$$ is added to indicade on which variable it acts. Now, what seperates us from the solution you are searching for, is just a partial integration to shift the differential to the first field in the integral. This should give you the wished result.