Discrete valuation ring extension such that $A[\pi]$ is not integrally closed

Let $B/A$ be a finite integral extension of discrete valuation rings of characteristic zero.

Question. Is there a uniformizer $\pi$ in $B$ such that $A[\pi]$ is a discrete valuation ring?

If not, can one give an explicit example? Note that to give such an example the extension of residue fields has to be INseparable. Is there an example with the extension of residue fields PURELY inseparable?

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Suppose $A$ is complete with imperfect residue field $k$ of characteristic $p>0$. Let $a\in A$ whose class in $k$ is not a $p$-th power. Let $t$ be a uniformizing element of $A$ and let $K=\mathrm{Frac}(A)$. Consider the field extension $$L=K[X]/(X^{p^2}-t^pa).$$ It has degree $p^2$. Let $B$ be the integral closure of $A$ in $L$. It is a DVR and is finite over $A$ because $A$ is complete. Let $\pi$ be the class of $X$ in $L$. Then $(\pi^p/t)^p=a$. So $e_{B/A}\ge p$ and because $a$ is not a $p$-th power in $k$, the degree of the residue extension is $\ge p$. Therefore $e_{B/A}=p$ and $\pi$ is a uniformizing element of $B$.
As $A[\pi]\simeq A[X]/(X^{p^2}-t^pa)$, it is not integrally closed by considering the element $\pi^p/t$.
In this example $B=A[\pi, \pi^p/t]$.