jmc

Let $P(x)=x^5+x^4+x^3+x^2+x$ and let $Q(x)$ be the quotient when $P(x)$ is divided by $x^3-4x$. Since we are dividing by a cubic, our remainder has degree at most $2$, meaning it is of the form $a{x}^2+bx+c$ for some constants $a$, $b$, and $c$. So we have $$P(x)=(x^3-4x)Q(x)+a{x}^2+bx+c.$$Since $x^3-4x=x(x^2-4)=x(x+2)(x-2)$, we can let $x=0,2$, or $-2$ to make $x^3-4x=0$. Doing each of them gives us the equations: $$\begin{array}{rl}0& =P(0)=c,\\ 62& =P(2)=4a+2b+c,\\ -22& =P(-2)=4a-2b+c.\end{array}$$Since $P(0)=0$, we know $c=0$. This gives us $4a+2b=62$ and $4a-2b=-22$. Solving these equations gives us $a=5$ and $b=21$.

Hence the remainder is $\boxed{5x^2+21x}$.