Let be $(R,\mathfrak{m},k)$ a Noetherian local ring. Denote by $Q$ the total ring of fractions of $R$. Let $I$ be an ideal of $R$ and $r \in I$ a $R$-regular element.
I have the following question: Given $y\in I$, is $\dfrac{y}{r}\in Q$ integral over $R$?
If the answer is no, under which conditions is the statement satisfied for all $y\in I$?
The origin of this question is because I am studying the proof of Theorem 3.3 of this paper (https://www.math.nagoya-u.ac.jp/~takahashi/tc9.pdf ) and trying to understand in the proof why $L:=\dfrac{I}{r} \subset \overline{R}$ ?
The fact that $\frac{I}{r} \subseteq \overline{R}$ is clearly equivalent to $I \subseteq r \overline{R}$ and to $I \subseteq r \overline{R} \cap R=\overline{(r)}$. Moreover, the containment $I \subseteq \overline{(r)}$ holds if and only if $(r)$ is a reduction of $I$ (see, for instance, Corollary of 1.2.5 of this book).
Therefore, in general the answer to your first question is no. For a concrete counterexample consider the integral domain $R=\mathbb{Q}[[t^2,t^3]]$, $I=(t^2,t^3)$, and $r=t^3 \in I$. In this case $\frac{t^2}{t^3}=t^{-1} \notin \overline{R}=\mathbb Q[[t]]$. On the other hand, if you choose $r=t^2$, you easily get $\frac{I}{t^2} \subseteq \overline{R}$; indeed, $(t^2)$ is a minimal reduction of $I$.