If the numerator degree is at least the denominator degree, divide first.
\(\dfrac{\mathrm{P}(x)}{\mathrm{D}(x)}=\mathrm{Q}(x)+\dfrac{\mathrm{R}(x)}{\mathrm{D}(x)}\)
Decompose the proper fraction \(\frac{R}{D}\), where \(R=0\) or \(\deg R<\deg D\). All identities require \(\mathrm{D}(x)\ne0\).
Distinct linear factors
\(\frac{px+q}{\left( ax+b \right)\left( cx+d \right)}=\frac{A}{ax+b}+\frac{B}{cx+d}\)
Repeated linear factor
\(\frac{p{{x}^{2}}+qx+r}{\left( ax+b \right){{\left( cx+d \right)}^{2}}}=\frac{A}{\left( ax+b \right)}+\frac{B}{\left( cx+d \right)}+\frac{C}{{{\left( cx+d \right)}^{2}}}\)
Irreducible quadratic factor
\(\frac{p{{x}^{2}}+qx+r}{\left( ax+b \right)\left( {{x}^{2}}+{{c}^{2}} \right)}=\frac{A}{\left( ax+b \right)}+\frac{Bx+C}{\left( {{x}^{2}}+{{c}^{2}} \right)}\)
Here \(a,c\ne0\); in the first two forms, \(ad-bc\ne0\). The quadratic \(x^2+c^2\) has no real roots.
Multiply through by the denominator. Substitute convenient values and compare coefficients to find the constants.
Include every power of a repeated factor, and restore any polynomial quotient.
Example
Express \(\dfrac{5x+1}{(x-1)(x+2)}\) in partial fractions.
\(\begin{aligned} \dfrac{5x+1}{(x-1)(x+2)}&\equiv\dfrac{A}{x-1}+\dfrac{B}{x+2} \\ 5x+1&\equiv A(x+2)+B(x-1) \end{aligned}\)
\(x=1:\ 6=3A,\ A=2;\quad x=-2:\ -9=-3B,\ B=3\)
\(\dfrac{5x+1}{(x-1)(x+2)}=\dfrac{2}{x-1}+\dfrac{3}{x+2}\)
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