jmc

Since our divisor $(x^2-4)(x+1)$ has degree $3$, our remainder must have degree at most $2$. In other words, our remainder is of the form $a{x}^2+bx+c$ for some constants $a$, $b$, and $c$. Let the quotient of the division be $q(x)$. Then $$x^5-x^4-x^3+x^2+x=(x^2-4)(x+1)q(x)+a{x}^2+bx+c$$We can see that our divisor $(x^2-4)(x+1)$ has roots $x=2$, $x=-2,$ and $x=-1$. Plugging in these roots gives us equations: For $x=2$ we have $32-16-8+4+2=0+4a+2b+c$ which gives us $$4a+2b+c=14.$$For $x=-2$ we have $-32-16+8+4-2=0+4a-2b+c$ which gives us $$4a-2b+c=-38.$$For $x=-1$ we have $-1-1+1+1-1=0+a-b+c$ which gives us $$a-b+c=1.$$Solving these three equations gives us $a=-8$, $b=13,$ and $c=20$.

So our remainder is $\boxed{-8x^2+13x+20}$.