Is $\int V^2e^{-V}d\text{vol}<\infty$ on a manifold such that $\text{Ric}+\text{Hess(V)}>0$?

Question

Suppose we have a measure $\mu(dx)= e^{-V(x)}d\text{vol}$ on a Riemannian manifold M with $V\in C^2(M)$, $\int e^{-V}d\text{vol} = 1$ and $\text{Ric}+\text{Hess(V)}>0$. Is it then true that $\int V^2e^{-V}d\text{vol}<\infty$? This holds on $\mathbb{R}^n$ and of course on any compact manifold, which tempts me to believe that it is always true, but I do not know for sure.