The answer is classical and negative. It is a particular instance of Thm X.11 of Reed-Simon here.

Let $V(r)$ be a continous symmetric potential on $\mathbb{R} \setminus \{0\}$. If \begin{equation} V(x) \geq \frac{3}{4x^2} \end{equation} Then $-\partial_x^2 +V$ it is essentially self-adjoint. On the other hand, if for some $\varepsilon >0$ \begin{equation} 0\leq V(x) \leq \left(\frac{3}{4} - \varepsilon\right)\frac{1}{x^2} \end{equation} Then $-\partial_x^2 +V$ is not essentially self-adjoint

The answer is classical and negative. It is a particular instance of Thm X.11 of Reed-Simon here.

Let $V(x)$ be a continous potential on $(0,+\infty)$. If \begin{equation} V(x) \geq \frac{3}{4x^2} \end{equation} Then $-\partial_x^2 +V$ it is essentially self-adjoint. On the other hand, if for some $\varepsilon >0$ \begin{equation} 0\leq V(x) \leq \left(\frac{3}{4} - \varepsilon\right)\frac{1}{x^2} \end{equation} Then $-\partial_x^2 +V$ is not essentially self-adjoint