If $\frac{1}{x^4(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+1}$ then $A+C+E=$

$1=A.x^3(x+1)+B{x}^2(x+1)+Cx(x+1)+D(x+1)+E.x^4$

Put $x=-1$, then $E=1$

Put $x=0$, then $D=1$

By comparing coefficients of $x^3,A+B=0\to \left(1\right)$

By comparing coefficients of $x,C+D=0\Rightarrow C=-1$

By comparing coefficients of $x^2,B+C=0\Rightarrow B=1$

$\left(i\right)\Rightarrow A+C=0$

$\Rightarrow A+B=0$

$\Rightarrow A=-1$

$\therefore A+C+E=-1-1+1=-1$