Partial fractions

Splitting a single algebraic fraction back into a sum of simpler ones — including the extra term a repeated linear factor needs.

Pure · Algebra & functions

Partial fractions

Splitting a single algebraic fraction back into a sum of simpler ones — including the extra term a repeated linear factor needs.

Why it works

Adding fractions combines them over a common denominator; partial fractions runs that in reverse, breaking one fraction into a sum whose denominators are the factors of the original. It's the key tool for integrating or expanding rational functions later, so it's worth doing fluently.

Distinct linear factors. Each factor of the denominator gets its own fraction with an unknown constant on top:

\frac{5x + 1}{(x - 1)(x + 2)} \equiv \frac{A}{x - 1} + \frac{B}{x + 2}.

Multiply through by the denominator:

5x + 1 \equiv A(x + 2) + B(x - 1)

. Now choose

x

to knock out a bracket (the "cover-up" idea):

$x = 1$ : $6 = 3A \Rightarrow A = 2$ ,
$x = -2$ : $-9 = -3B \Rightarrow B = 3$ .

\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{2}{x - 1} + \dfrac{3}{x + 2}

.

Repeated linear factors. A factor like

(x + 1)^2

needs two terms — one for each power:

\frac{\dots}{x\,(x + 1)^2} \equiv \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}.

Forgetting the

\dfrac{B}{x+1}

term is the classic slip. After substituting the "nice" values to find

A

and

C

, the middle constant

B

is found by equating coefficients (e.g. of

x^2

) or substituting one more value.

Always factorise the denominator first, and remember every top is a constant (for linear factors) — a value to be found, not left blank.