Partial Fractions With Repeated Quadratics

Question

I'm told that given a function $f(x)=\frac{P(x)}{Q(x)}$, if $\deg(P)>\deg(Q)$ then $f$ is improper, which makes sense when I think of real numbers like $5/2$. And in this case we would have to do long division to ensure that the degree of the denominator is greater. Something of the form $S(x)+\cfrac{R(x)}{Q(x)}$.

My book eventually gives the example problem of case 4, when $Q(x)$ contains repeated quadratics $$\cfrac{x^3+x^2+1}{x(x-1)(x^2+x+1)(x^2+1)^3}$$

So now I'm thinking since the numerator cannot be factored and the leading degree is greater than $x$, then the first term of the partial fraction should be $\cfrac{Ax+B}{x}$, likewise since it's also greater than $(x-1)$ then the second term is $\cfrac{Cx+D}{x-1}$, then $\cfrac{Ex+F}{(x^2+x+1)}+\cfrac{G}{(x^2+1)}+\cfrac{H}{(x^2+1)^2}+\cfrac{I}{(x^2+1)^3}$

Instead, my textbooks solution obtains $$\cfrac{A}{x}+\cfrac{B}{x-1}+\cfrac{Cx+D}{x^2+x+1}+\cfrac{Ex+F}{x^2+1}+\cfrac{Gx+H}{(x^2+1)^2}+\cfrac{Ix+J}{(x^2+1)^3}$$

Could anyone explain what part of my interpretation here is wrong?

Answer 1

Let $\cfrac{P(x)}{Q(x)}=\cfrac{x^3+x^2+1}{x(x-1)(x^2+x+1)(x^2+1)^3}.$

This rational function is a proper fraction because $\deg(P)=3<10=\deg(Q)$,

and its denominator has been fully factorised into linear and irreducible quadratic factors;

so we can proceed to decompose it into partial fractions according to these rules:

1. To each linear factor $(ax+b)^n$, there corresponds partial fractions of the form $\cfrac{A_1}{(ax+b)},\cfrac{A_2}{(ax+b)^2},\ldots,\cfrac{A_n}{(ax+b)^n}$.
2. To each irreducible quadratic factor $(ax^2+bx+c)^n$, there corresponds partial fractions of the form $\cfrac{A_1x+B_1}{(ax^2+bx+c)},\cfrac{A_2x+B_2}{(ax^2+bx+c)^2},\ldots,\cfrac{A_nx+B_n}{(ax^2+bx+c)^n}$.

By the Cover-up Rule, $\cfrac{P(x)}{Q(x)}=\cfrac{-1}{x}+\cfrac{1}{8(x-1)}+\cfrac{Cx+D}{x^2+x+1}+\cfrac{Ex+F}{x^2+1}+\cfrac{Gx+H}{(x^2+1)^2}+\cfrac{Ix+J}{(x^2+1)^3}$.

Equating coefficients gives $C=-1,D=-1,E=\frac{15}{8},F=\frac{-1}{8},G=\frac{3}{4},H=\frac{3}{4},I=\frac{-1}{2},J=\frac{1}{2}$.

Therefore $$\cfrac{P(x)}{Q(x)}=\cfrac{-1}{x}+\cfrac{1}{8(x-1)}-\cfrac{x+1}{x^2+x+1}+\cfrac{15x-1}{8(x^2+1)}+\cfrac{3(x+1)}{4(x^2+1)^2}+\cfrac{1-x}{2(x^2+1)^3}.$$