Partial Fractions

Main idea

Partial fractions are fraction addition run backwards: you split one complicated fraction into a sum of simple ones, one for each factor of the denominator. Simple pieces are much easier to work with later.

\frac{px + q}{(x - a)(x - b)} = \frac{\htmlData{sym=0}{{A}}}{x - a} + \frac{\htmlData{sym=1}{{B}}}{x - b}

What the symbols mean

$A$
First numerator
An unknown constant, found by matching.
$B$
Second numerator
Another unknown constant.

Explanation

Partial fractions are simply what you get when you undo adding fractions. If you add $\frac{2}{x - 3}$ and $\frac{-1}{x + 2}$ you get one fraction, $\frac{x + 7}{(x - 3)(x + 2)}$ ; partial fractions take you from that back to the two simple pieces, the "parts" of the fraction. In the formula, the left side is the combined fraction, and $A$ and $B$ (unknown numbers) are the tops of the simple pieces, one per factor of the bottom. All the work is finding $A$ and $B$ . It sounds odd to make one fraction into two, but you'll lean on this heavily in calculus.

Why the pieces look like this. Think about how you'd add $\frac{A}{x - 3} + \frac{B}{x + 2}$ : the common denominator is $(x - 3)(x + 2)$ . So any fraction with that denominator ought to be splittable back into pieces over $x - 3$ and $x + 2$ . The denominator's factors tell you exactly which pieces to expect.

Which pieces to use, for each factor of the denominator. This needs the top's degree to be lower than the bottom's; if it isn't, divide first, like turning $\frac73$ into $2 + \frac13$ .

Factor	Pieces
$x - a$	$\dfrac{A}{x - a}$
$(x - a)^2$	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$
$x^2 + c$ (doesn't factor)	$\dfrac{Ax + B}{x^2 + c}$

A repeated factor like $(x - a)^2$ needs both pieces. A single $\frac{B}{(x - a)^2}$ only has a constant on top, but the top over $(x - a)^2$ could be any linear expression. Together, $\frac{A}{x - a} + \frac{B}{(x - a)^2} = \frac{A(x - a) + B}{(x - a)^2}$ , which can match any linear top. A quadratic that won't factor gets a linear top, $Ax + B$ , since one degree less than the bottom is the most the top can be.

Finding the constants. Multiply both sides by the denominator to clear all the fractions. Then either plug in convenient values of $x$ (the zeros of the factors, which wipe out all but one term) or match the coefficients of each power of $x$ on both sides.

The cover-up shortcut for a distinct linear factor $x - a$ : cover that factor in the original fraction and evaluate what's left at $x = a$ . It's the plug-in method done in your head.

See it

Here's (x + 7)/((x − 3)(x + 2)) next to its two pieces, 2/(x − 3) and −1/(x + 2), drawn dashed. Notice each piece is responsible for one of the vertical asymptotes, at 3 and −2. Slide x and you'll see the two dashed heights always add up to the solid one. That splitting is what later makes these easy to integrate.

Worked examples

Example 1

Decompose $\dfrac{x + 7}{x^2 - x - 6}$ .

1
Factor the bottom: $x^2 - x - 6 = (x - 3)(x + 2)$ . Set up $\frac{A}{x - 3} + \frac{B}{x + 2}$ and multiply through by the denominator: $x + 7 = A(x + 2) + B(x - 3)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Decompose $\dfrac{x + 7}{(x - 3)(x + 2)}$ . Enter $A$ , $B$ .

Problem 2+5 XP

Decompose $\dfrac{3}{x^2 - 1} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$ . Enter $A$ , $B$ .

Problem 3+5 XP

Decompose $\dfrac{2x + 3}{(x - 1)^2}$ . Enter the sum of partial fractions.

Problem 4+5 XP

Which form fits $\dfrac{x^2 + 1}{x(x^2 + 4)}$ ?

Problem 5+5 XP

Find $A$ , $B$ , $C$ in $\dfrac{x^2 + 1}{x(x^2 + 4)} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 4}$ .

Problem 6+5 XP

Decompose $\dfrac{1}{n(n + 1)} = \dfrac An + \dfrac{B}{n + 1}$ . Enter $A$ , $B$ .

Problem 7+5 XP

Use that to add $\dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} + \cdots + \dfrac{1}{99\cdot100}$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Decompose $\dfrac{6x^2 + 6x - 6}{x^3 + x^2 - 2x} = \dfrac Ax + \dfrac{B}{x - 1} + \dfrac{C}{x + 2}$ . Enter $A$ , $B$ , $C$ .

Where it's used

Partial fractions make hard integrals easy in Calculus II and turn Laplace transforms back into time signals in differential equations and control engineering.

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