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

If we resolve the rational fraction $\frac{1}{(1-2x)^2(1-3x)}$ into partial fractions of the form $\frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$,then what is $\min \{A, B, C\} = $

$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A + 2B + C = ?$

The coefficient of $x^3$ in the expansion of $\frac{x^4+1}{(x^2+1)(x-1)}$ when it is expressed in terms of positive integral powers of $x$,is

If $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$,then $k =$

Let $x$ be a real number and $-2 < x < 2$. When $\frac{x+1}{(x+3)(x-2)}$ is expanded in powers of $x$,then the coefficient of $x^3$ is