Problem on understanding why the ring $R[x]$ is integrally closed provided $R$ is integrally closed I was trying to understand the discussion here about showing that $R[x]$ is integrally closed provided that $R$ is integrally closed; Badam Baplan provides a potential solution and at some step, he asserts that if $f \in K[x]$ is integral over $R[x]$ (where $K$ is the quotient field of $R$) with constant term $f_0,$ then $f-f_0$ is also integral over $R[x]$ (this is ok since the set of integrally closed elements of a ring form a ring and $f_0$ is clearly integral over $R[x]$). But then he asserts that $\frac{f-f_0}{x}$ is integral over $R[x],$ which is the part where I'm lost at.
He also gives an argument about why this is true but it is pretty unclear to me. Can someone explain to me (considering I'm just a beginner on commutative algebra) why is it true that $\frac{f-f_0}{x}$ is integral over $R[x]$? Or maybe provide another 'elementary' way to show that $R[x]$ is integrally closed provided $R$ is integrally closed?
Thanks in advance.

Note: as mentioned in a comment in the linked post, the polynomial is not necessarily monic, so this doesn't actually show $(a'- a )/x$ is integral.
I'll use the notation in the given post : $a' = a - a_0$ is integral over $R[x]$ , so there is some monic polynomial with coefficients in $R[x]$, say $p(y) \in R[x][y]$ such that $p(a') = 0$. Write $p(a') = \sum_{i=0}^k d_i(x) a'^i  = 0$ in $R[x]$
$a'$ has no constant term, so reducing the above equation mod $x$, we get that $p(a') = d_0(x) = 0 \mod x$. Therefore, $d_0$ must also not have a constant term.
Therefore, write $a' = x*b, d_0 = x*c$ for $b, c \in R[x]$. We have
$\begin{align}
\sum_{i=0}^k d_i(x) a'^i\\
 =  d_0(x) + \sum_{i=1}^k d_i(x) a'^i \\
= xc + \sum_{i=1}^k d_i(x) (bx)^i \\
 = x \left(c + \sum_{i=1}^k d_i(x) b^ix^{i-1}\right)
\end{align}
$
Therefore, $b = a' / x = (a - a_0)/x$ is the root of a polynomial $c + \sum_{i=1}^k d_i(x) x^{i-1} y^i \in R[x][y]$
