Continuation of the previous question: Partial Fraction Decomposition: Heaviside's Cover-up methods for repeated roots
Note: We try to find a method that's almost identical to Heaviside cover-up method and limit method, not the common method like cross-multiplication.
Let's talk about another function: $$f\left ( x \right )=\frac{x^{5}-x^{4}+x^{3}-2x^{2}}{x^{6}+3x^{4}+3x^{2}+1}=\frac{x^{5}-x^{4}+x^{3}-2x^{2}}{\left ( x^{2}+1 \right )^{3}}$$
This function can be rewritten as: $$f\left ( x \right )=\frac{a_{1}x+b_{1}}{x^{2}+1}+\frac{a_{2}x+b_{2}}{\left ( x^{2}+1 \right )^{2}}+\frac{a_{3}x+b_{3}}{\left ( x^{2}+1 \right )^{3}}$$
In the previous situation where people and I uncovered, the repeated terms are all linear, and obtaining the coefficient of those can be done by turning $f(x)(x-x_{0})^{n}$ into a Taylor Series at $x=x_{0}$. Now, it is even harder to obtain the coefficients if we account for repeated irreducible quadratics (apart from the highest order, likewise), because in this situation you can't just rewrite $f(x)(x^2+1)^3$ into an infinite series. So, my question is how to obtain those coefficients?
One of my approaches is to split $x^2+1$ into $(x-i)(x+i)$ and have the partial fraction decomposition of smaller terms: $$f\left ( x \right )=\frac{m_{1}}{x-i}+\frac{m_{2}}{\left ( x-i \right )^{2}}+\frac{m_{3}}{\left ( x-i \right )^{3}}+\frac{n_{1}}{x+i}+\frac{n_{2}}{\left ( x+i \right )^{2}}+\frac{n_{3}}{\left ( x+i \right )^{3}}$$ Then I find $m_1$ and $m_2$ by rewriting $f(x)(x-i)^3$ into a Taylor Series at $x=i$, and it goes the same with $f(x)(x+i)^3$.
I think this approach isn't quite friendly to people who haven't even got into complex numbers. Even so, if the factor gets more complex like $x^2+2x+3$, it will be even harder to work. I wonder if there is a better approach without accounting for complex numbers.
