May assume field $k=\mathbb{C}$.
Let $X$ be an affine variety and $G$ be a reductive group (may assume connected). 


Is the ring of invariants $k[X]^G$ integral closed in $k[X]$?


The claim may not true in general, probably we may assume $X$ is normal and the affine quotient $X/G$ also normal.


Any reference?



Is $k[X]^G$ integral closed in $k[X]$.
Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$.


In fact, for $G$ not connected with $G\_0$ the connected component of the identity and $G/G\_0$ finite, the integral closure of $k[X]^G$ in $k[X]$ is $k[X]^{G\_0}$ because any element in $k[X]^{G\_0}$ has a finite orbit and the coefficients of the monic polynomial that vanishes exactly on the orbit are $G$-invariant.


