If $\frac{x^4}{(x^2+1)(x-2)}=f(x)+\frac{Ax+B}{x^2+1}+\frac{C}{x-2}$, then $f(14)+2A-B=$ (in $C$)

If $\frac{(x - a)(x - b)}{(x - c)(x - d)} = \frac{A}{x - c} - \frac{B}{x - d} + C$,then $C = $

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

If $\frac{3}{(x-1)(x^2+x+1)} = \frac{1}{x-1} - \frac{x+2}{x^2+x+1} = f_1(x) - f_2(x)$ and $\frac{x+1}{(x-1)^2(x^2+x+1)} = A f_1(x) + (B + \frac{D}{x-1}) f_2(x) + \frac{C}{(x-1)^2}$,then find $A+B+C+D$.

If the equivalent partial fraction of $\frac{x^3}{(2x-1)(x+2)(x-3)}$ is given by $A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,then the value of $C$ is

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