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

If $\frac{3x^2+x+1}{(x-1)^4} = \frac{a}{(x-1)} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4}$,then $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is equal to

If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$, then $a-b+c=$

If $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$,then $A - B - C = $

If the partial fraction decomposition of $\frac{x^2+1}{x^3+3x^2+3x+2}$ is $\frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$,then find the value of $A-B+C$. Note: The original expression provided in the prompt was corrected to the standard form $\frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$.

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C= \end{aligned}$