Let $A$ be a domain with field of fractions $F$, integrally closed in $F$, and let $\alpha$ be algebraic over $F$.
Then $\alpha$ is integral over $A$ if and only if its monic irreducible polynomial over $F$ has coefficients in $A$.
Moreover, if $\alpha\neq 0$ and $f(x)$ is the monic irreducible polynomial of $\alpha$, then the monic irreducible polynomial of $\frac{1}{\alpha}$ is $x^nf(\frac{1}{x})$, where $n=\deg(f)$.
In particular, $\alpha$ and $\frac{1}{\alpha}$ are both integral over $A$ if and only if the monic irreducible polynomial of $\alpha$ has coefficients in $A$ and its constant term is a unit of $A$.