Partial fractions

What you’ll learn

• Decompose rational functions with distinct or repeated linear factors.
• Divide improper rational expressions before decomposition.

Understand the idea

Partial fractions reverse the process of adding fractions. Each distinct linear factor in a denominator contributes a constant numerator over that factor. Multiply by the complete denominator and compare coefficients or substitute convenient values to determine the constants.

A repeated factor needs a term for every power up to its multiplicity. For example, a squared factor requires both $A/(x-a)$ and $B/(x-a)^2$. If the numerator has degree at least as large as the denominator, first divide to get a polynomial plus a proper fraction.

The decomposition preserves the original excluded values of $x$. It can simplify integration, differentiation or a series expansion. Recombine the resulting fractions as a quick independent check on the constants and signs.

Key formulae

$$\frac{px+q}{(x-a)(x-b)}=\frac A{x-a}+\frac B{x-b}\quad(a\ne b)$$

$$\frac{P(x)}{(x-a)(x-b)^2}=\frac A{x-a}+\frac B{x-b}+\frac C{(x-b)^2}\quad(\deg P<3)$$

$$\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)},\quad\deg R<\deg Q$$

Worked example

Decompose $\dfrac{5x+1}{(x-1)(x+2)}$ into partial fractions.

1. Write $A/(x-1)+B/(x+2)$. Multiply through to get $5x+1=A(x+2)+B(x-1)$.
2. Set $x=1$: $6=3A$, so $A=2$.
3. Set $x=-2$: $-9=-3B$, so $B=3$.

Answer: $\dfrac2{x-1}+\dfrac3{x+2}$, for $x\ne1,-2$.