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.

Let $\bar{R} = R / rR$. Then $R[x] / rR[x] \cong \bar{R}[X]$ and the image $\bar{S}$ of $S$ in $\bar{R}[\bar{x}]$ is contained in $\bar{R}$ by assumption.
Suppose $x$ is integral over $S$. Then $X$ is integral over $\bar{S}$. Hence $X$ is integral over $\bar{R}$. The indeterminate $X$ is integral over $\bar{R}$ inside the polynomial ring $\bar{R}[X]$ only when $\bar{R}$ is the zero ring. So $rR = R$ and $r$ is a unit in $R$, a contradiction.
