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

The partial fraction decomposition of $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3}$ is:

If $\frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx-3}{x^2+2}$,then $A+B+C=$

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \ldots + \frac{A_n}{x+n}$. For $0 \leq r \leq n$,$A_r$ is equal to:

Let $\frac{1}{(x^2-3)^2} = \frac{A_1}{x-\sqrt{3}} + \frac{A_2}{(x-\sqrt{3})^2} + \frac{A_3}{x+\sqrt{3}} + \frac{A_4}{(x+\sqrt{3})^2}$. Then,consider the following statements:
$(i)$ All the $A_i$'s are not distinct
(ii) There exists a pair,$A_p$ and $A_q$ such that $A_p^2 = A_q^2$ $(p \neq q)$
(iii) $\sum_{i=1}^4 A_i = \frac{1}{6}$
(iv) $\sum_{i=1}^4 A_i = 1$
Which one of the following is true?

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