An element $\alpha\in L/K$ is integral over A if its minimal polynomial over $K$ has coefficients in $A$. I suspect it is a very trivial question, it is marked as such in algebraic number theory books by Neukirch and Lee, and also in lectures I'm attending. But I can't wrap my head around it.
So, $L/K$ is an algebraic field extension, $A\subseteq K$ - an integrally closed integral domain, such that $Quot(A) = K$.
In the books, "iff" version of this statement is proved, but actually only the opposite direction is proved, and the one I'm asking about is stated as trivial.

If $f = a_n X^n + \dots + a_0$, $a_i\in A,$ is the minimal polynomial of $\alpha$, how do we get rid of $a_n$?

Edit: I forgot that minimal polynomial is defined to be monic. Can be closed.
One direction is trivial. I'll show the non-trivial one, that is, if $\alpha \in L$ is integral over $A$, then the minimal polynomial of $\alpha$ over $K$ has its coefficients in $A$.

Since $\alpha$ is integral over $A$, there is $p\in A[x]$, a  monic polynomial, for which $p(\alpha)=0$. 

Of course $\alpha $ is algebraic over $K$, so let $q\in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Then $q$ divides $p$ in $K[x]$. Hence the roots of $q$ are all integral over $A$ (because these roots are just some of the roots of $p\in A[x]$).

Write $$q(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_0=\prod_i (x-\alpha_i) \in K[x].$$
But you know we can express the coefficients $a_i$ in terms of symmetric polyomials in the roots $\alpha_i$'s. Hence, it follows that each $a_i$ is integral over $A$. But as $A$ is integrally closed, it follows $a_i \in A$ for all $i$.
"$\alpha$ integral over $A$" : $\alpha$ is a root of some polynomial $\in A[x]_{monic}$ , equivalently $A[\alpha]$ is a finitely generated $A$-module, equivalently $\alpha \in End(M)$ where $M$ is a finitely generated free $A$-module (since then $\alpha$ is a root of the characteristic polynomial of the endomorphism $\in End(M)$ of the multiplication by $\alpha$).
If $A$ is integrally closed in $Frac(A)$ then this is the same as "$f\in K[x]_{monic}$, the minimal polynomial of $\alpha$, is in $A[x]$". One direction is obvious, the other is that $\alpha$ is a root of $g=\det(\alpha-xI)\in A[x]_{monic}$, since every root of $f$ is a root of $g$ we have $f^m=g\in A[x]$ which implies that the coefficients of $f$ are integral over $A$. Since $A$ is integrally closed in $K$ it means $f \in A[x]$.
Thus in that case it is easy to see that if $\alpha,\beta$ are integral then $A[\alpha,\beta]$ is a finitely generated $A$-module so all its elements are integral and  hence the integral closure of $A$ is a ring.