We have shown that, for static fields,
$$
\boldsymbol{E}=\frac{1}{4 \pi \epsilon_{0}} \int_{v} \cdot \frac{\rho}{r^{2}} \hat{\boldsymbol{r}} d v^{\prime}=-\frac{1}{4 \pi \epsilon_{0}} \int_{v^{\prime}} \frac{\boldsymbol{\nabla} \rho}{r} d v^{\prime}
$$
Show that, as a consequence,
$$
\int_{v^{\prime}} \boldsymbol{\nabla}^{\prime} \frac{\rho}{r} d v^{\prime}=0
$$