$\nabla\cdot\left(\hat{\mathbf{r}}/r^{2}\right) =0$ where $r$ is in spherical coordinates and represents the distance from the origin.
In the Griffith' Electrodynamics pg 46, first line it is stated that the volume integral
$$ {\Large\int_V} {\nabla \cdot ( \frac{\hat{\mathbf{r}}}{r^2}) \, d\tau}=0$$
I am not able to wrap my head around this statement. How can volume integral of zero function be zero? Shouldn't it be some constant $C$ or any other non-zero expression?
I am looking for a mathematical answer by solving the integral.
$$ {\Large\int_{r=0}^1\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}} {\nabla \cdot ( \frac{ \hat{\mathbf{r}} }{r^2}) r^2 sin\theta\, dr d\theta d\phi}$$
I am stuck at the part of $$ {\Large\int_{0}^1} {\nabla \cdot ( \frac{ \hat{\mathbf{r}} }{r^2}) r^2 dr}$$ and don't know how to proceed further.
Edit: Included $\hat{\mathbf{r}}$
Edit 2: I got an idea of correct answer from the answer of @kandb and comment of @ninad_Munshi but I would prefer an answer that will be able to help me solve the last integral
The volume integral is not zero--Griffiths is being didactic in saying the volume integral is zero "if we are really to believe Eq. 1.84." The tricky thing is that
\begin{equation*} \nabla \cdot \left(\frac{\hat{\mathbf{r}}}{r^{2}}\right) = \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\cdot\frac{1}{r^{2}}\right) = 0 \end{equation*}
everywhere except at the origin, where it is unbounded. The correct form for the divergence is
\begin{equation*} \nabla \cdot \left(\frac{\hat{\mathbf{r}}}{r^{2}}\right) = 4\pi \delta^{3}(\mathbf{r}). \end{equation*}
(You can see that this must be the case using the divergence theorem to convert the volume integral to a surface integral.) Then you have that
\begin{equation*} \int{\nabla \cdot \left(\frac{\hat{\mathbf{r}}}{r^{2}}\right)\mathrm{d}V} = \int{4\pi \delta^{3}{(\mathbf{r})}\mathrm{d}V} = 4\pi. \end{equation*}
Edit: To evaluate your integral directly, you can use a form of the Dirac delta function for spherical coordinates in the case of spherical symmetry (you are integrating over the unit sphere centered at the origin, which is of course spherically symmetric; the form is different if you only have azimuthal symmetry or if you don't have any symmetry):
\begin{equation*} \delta(\mathbf{r}) = \frac{1}{4\pi r^{2}}\delta(r). \end{equation*}
Then, your integral becomes
\begin{align*} \int{\nabla \cdot \left(\frac{\hat{\mathbf{r}}}{r^{2}}\right)\mathrm{d}V} &= \int_{0}^{2\pi}{\int_{0}^{\pi}{\int_{0}^{1}{4\pi \delta(\mathbf{r})r^{2}\sin{(\theta)}\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi}}}\\ &= 16\pi^{2}\int_{0}^{1}{\frac{1}{4\pi r^{2}}\delta(r)r^{2}\mathrm{d}r}\\ &= 4\pi. \end{align*}