Consider $R[x]$ and let $S$ be the subring generated by $rx$, where $r \in R$ is some non-invertible element. Then $x$ is not integral over $S$ 
Consider $R[x]$ and let $S$ be the subring generated by $rx$, where $r \in R$ is some non-invertible element. Then I want to show that $x$ is not integral over $S$

I'm not seeing why this is the case.
Note : Here R is a commutative ring.

Assume that $x$ is integral over $S$. Then there is a polynomial $p(X)=x^n+a_1X^{n-1}+...a_n$, with $a_i$ in $S$ such that $p(x)=0$.
But since $a_i$ are in $S$, they are of the form $q_i(rX)$ for some polynomials $q_i$ with integer coefficients. 
We get $0=p(x)=x^n+q_1(rx)x^{n-1}+-...+q_{n}(rx)$. The latter is a polynomial in $x$. For it to be zero, the leading term must be zero. But the $x^n$ term is a multiple of $r$, say $rA$ plus $1$. But if $rA+1=0$, then $(-A)r=1$, which implies $r$ is invertible.
