jmc

Since $x^3-x=x(x^2-1)=x(x+1)(x-1)$ has degree $3$, we know that the remainder is of the form $a{x}^2+bx+c$ for some constants $a$, $b$, and $c$. Let the quotient be $q(x)$. Then, $$x^{18}+x^{13}+x^7+x^4+x=x(x+1)(x-1)q(x)+a{x}^2+bx+c.$$If we plug in $x=0$, we get $c=0$. If we plug in $x=1$, we get $5=a+b$. And if we plug in $x=-1$ we get $-1=a-b$. Solving these two equations together gives us $a=2$ and $b=3$ which means the remainder is $\boxed{2x^2+3x}$.