electrostatics - Why does Jackson ignore the surface term in this integration by parts, in the electric potential of a polarized dielectric?

In Jackson's electrodynamics Page (153), it is given that the total potential is the sum of potential due to free charge and potential due to dipole. Therefore, \begin{align*} \phi(x) & = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{x}')}{|\vec{x} - \vec{x}'|}d^3x' + \frac{1}{4\pi\varepsilon_0}\int{\vec{P}(\vec{x}')}\cdot\vec{\nabla}'\frac{1}{|\vec{x} - \vec{x}'|}d^3x' \\ & = \frac{1}{4\pi\varepsilon_0}\left[\int\frac{\rho(\vec{x}')}{|\vec{x} - \vec{x}'|}d^3x' + \int\vec{\nabla}'\cdot\left[\frac{{\vec{P}(\vec{x}')}}{|\vec{x} - \vec{x}'|} \right]d^3x' - \int\frac{\vec{\nabla}'\cdot \vec{P}(\vec{x}')}{|\vec{x} - \vec{x}'|}d^3x' \right] \end{align*} Using the Gauss's divergence theorem, the 2nd term becomes \begin{equation*} \int\vec{\nabla}'\cdot\left[\frac{{\vec{P}(\vec{x}')}}{|\vec{x} - \vec{x}'|} \right]d^3x' = \int\frac{\vec{P}(\vec{x}')\cdot \vec{ds}'}{|\vec{x} - \vec{x}'|} \end{equation*} This term has been neglected while reaching Eq. (4.32) in the book. How we can do that?