If $\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$,then $A+B$ is equal to

$\frac{x^4}{x^3-3x+2}$ is a

The partial fraction of $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ is:

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

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