Let $A = \mathbb{C}[x^5, y^5, x^4y]$. I want to find its integral closure in $\mathrm{Frac}(A)$, i.e. the normalization of $A$.
As given in the comments, the answer is $\mathbb{C}[x^5, y^5, x^4y, x^3y^2, x^2y^3, xy^4, y^5]$, but I still don't know why.
$S^{G} =\mathbb{C}[x^5, y^5, x^4y, x^3y^2, x^2y^3, xy^4, y^5]$ is the ring of invariants of $S = \mathbb{C}[x,y]$ by cyclic group of order 5.
That means that $S$ is integral over $S^G$.
So, we have $(x^2y^3)^5=(x^5)^2\cdot(y^5)^3$. This means that $x^2y^3$ is integral over $A$
That means that the answer contains all monomials of degree 5, which is $S^G$, where $G = \mathbb{Z}/5\mathbb{Z}$.
Algebra $S^G$ is normal, because $S$ is a UFD hence normal.