Introduction to Partial Fractions

By the end of this lesson, you will be able to decompose proper rational functions into partial fractions for distinct linear, repeated linear, and irreducible quadratic factors in the denominator.

The Big Idea

Partial fraction decomposition reverses the process of adding rational expressions — breaking one complex fraction into a sum of simpler ones. This is essential for integration in calculus (every rational function is integrable by partial fractions) and for inverse Laplace transforms in engineering.

Setup Rules

A rational function $\dfrac{P(x)}{Q(x)}$ must be proper ($\deg P < \deg Q$). If not, divide first.

For each factor in $Q(x)$:

Factor in denominator	Partial fraction form
$(x - a)$ (distinct linear)	$\dfrac{A}{x-a}$
$(x - a)^2$ (repeated linear, degree 2)	$\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$
$(x - a)^n$ (repeated, degree $n$)	$\dfrac{A_1}{x-a} + \cdots + \dfrac{A_n}{(x-a)^n}$
$x^2 + bx + c$ (irreducible quadratic)	$\dfrac{Ax + B}{x^2+bx+c}$

Example 1 — Distinct Linear Factors

Decompose: $\dfrac{7x - 1}{(x-3)(x+2)}$

$$\frac{7x-1}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$$

Multiply both sides by $(x-3)(x+2)$:

$7x - 1 = A(x+2) + B(x-3)$

Method — strategic substitution:

$x = 3$: $21 - 1 = 5A \implies A = 4$

$x = -2$: $-14 - 1 = -5B \implies B = 3$

$$\frac{7x-1}{(x-3)(x+2)} = \frac{4}{x-3} + \frac{3}{x+2}$$

Example 2 — Repeated Linear Factor

Decompose: $\dfrac{5x+2}{x^2(x-1)}$

$$= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}$$

Multiply: $5x + 2 = Ax(x-1) + B(x-1) + Cx^2$

$x = 0$: $2 = -B \implies B = -2$

$x = 1$: $7 = C \implies C = 7$

Expand and compare $x^2$ coefficients: $0 = A + C = A + 7 \implies A = -7$

$$= \frac{-7}{x} + \frac{-2}{x^2} + \frac{7}{x-1}$$

Example 3 — Irreducible Quadratic

Decompose: $\dfrac{2x^2 + 3x - 1}{x(x^2+1)}$