Question

(A/x-1)+(B/(x-1)^2)+(Cx+D/(x^2+3))...what should be the numerator?

Answer 1

That depends on what this is equal to...
\[\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+3}=\cdots\]
Combining the fractions, you'd have
\[\frac{A(x-1)(x^2+3)+B(x^2+3)+(Cx+D)(x-1)^2}{(x-1)^2(x^2+3)}=\cdots\\
\frac{(A+C)x^3+(B-A-2C+D)x^2+(3A+C-2D)x+(D-3A+3B)}{(x-1)^2(x^2+3)}=\cdots\]
Let's say the right hand side's numerator is a general 3rd degree polynomial: \(ax^3+bx^2+cx+d\). Matching up the coefficients, your job would be to solve the system:
\[\begin{cases}
\begin{align*}A+C&=a\\B-A-2C+D&=b\\3A+C-2D&=c\\D-3A+3B&=d\end{align*}
\end{cases}\]