I think it's useful to expand slightly on the comments because I think the language here should be as clear as possible.

An integral of a function $f(x)$ over a domain $[a, b]$ will not converge if there exists $c\in [a, b]$ such that $f(c) = \infty$.

Your function certainly looks like that but the resolution of this problem is actually hidden in the $dV$. Typically, we treat the integration of the wave function as a separable problem, i.e, we integrate the radial and spherical components separately. When you look at the integral over the distance from the nucleus, you get a factor $4\pi r^{2}$ because you are integrating shells and $dV = 4\pi r^{2} dr$. So even though $\frac{1}{r}$ diverges at zero, $\frac{r^{2}}{r}$, while not defined at zero, does converge.

$$\lim_{r\rightarrow 0}\frac{r^{2}}{r} = 0$$

From this perspective, no piece of the integral is infinite, and the integral converges.

I think it's useful to expand slightly on the comments because I think the language here should be as clear as possible.

You're concerned about the convergence of the integral of a function when some piece of the function is unbounded, i.e., for $f(x)$ on domain $[a, b]$, $\exists c \in [a,b]$ such that $f(c) = \infty$. Frequently, integrals of this type do not converge.

Your function certainly looks like that but the resolution of this problem is actually hidden in the $dV$. Typically, we treat the integration of the wave function as a separable problem, i.e, we integrate the radial and spherical components separately. When you look at the integral over the distance from the nucleus, you get a factor $4\pi r^{2}$ because you are integrating shells and $dV = 4\pi r^{2} dr$. So even though $\frac{1}{r}$ diverges at zero, $\frac{r^{2}}{r}$, while not defined at zero, does converge.

$$\lim_{r\rightarrow 0}\frac{r^{2}}{r} = 0$$

From this perspective, no piece of the integral is infinite, and the integral converges.