You know how to add rational expressions: $\frac{2}{x - 1} + \frac{3}{x + 1} = \frac{5x - 1}{x^2 - 1}$. Partial fractions run that process backward, splitting one fraction into simple pieces. Calculus needs this to integrate rational functions, and finding the pieces is a small system of equations.
Key idea
For a proper fraction (numerator degree less than denominator degree), factor the denominator and write one piece per factor: $$\frac{A}{x - a} \text{ for each linear factor}, \quad \frac{A}{x - a} + \frac{B}{(x - a)^2} \text{ for a repeated factor}, \quad \frac{Bx + C}{x^2 + c} \text{ for an irreducible quadratic}.$$ Clear denominators, then find the constants by substituting convenient $x$-values or matching coefficients.
Why it works
When you add the pieces over the common denominator, the numerator you get must equal the original numerator for every $x$. Each piece contributes the right number of unknowns: a quadratic factor needs a numerator of degree up to one, $Bx + C$, because adding fractions can produce any numerator smaller in degree than the denominator. Substituting $x = a$ makes every term except one vanish, which is why roots are such convenient values to plug in.
Worked examples
Worked example: Distinct linear factors
Decompose $\dfrac{5x - 1}{(x - 1)(x + 1)}$.
- Write $\dfrac{A}{x - 1} + \dfrac{B}{x + 1}$ and clear denominators: $5x - 1 = A(x + 1) + B(x - 1)$.
- $x = 1$: $4 = 2A$, so $A = 2$.
- $x = -1$: $-6 = -2B$, so $B = 3$.
Answer: $\frac{2}{x - 1} + \frac{3}{x + 1}$
Worked example: A repeated factor
Decompose $\dfrac{3x + 5}{(x + 1)^2}$.
- Write $\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2}$: $3x + 5 = A(x + 1) + B$.
- Match the $x$ terms: $A = 3$. Then $x = -1$: $2 = B$.
Answer: $\frac{3}{x + 1} + \frac{2}{(x + 1)^2}$
Worked example: An irreducible quadratic
Decompose $\dfrac{3x^2 + 2x + 5}{(x - 1)(x^2 + 4)}$.
- Write $\dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 4}$: $$3x^2 + 2x + 5 = A(x^2 + 4) + (Bx + C)(x - 1).$$
- $x = 1$: $10 = 5A$, so $A = 2$.
- Match $x^2$ terms: $3 = A + B$, so $B = 1$. Match constants: $5 = 4A - C$, so $C = 3$.
Answer: $\frac{2}{x - 1} + \frac{x + 3}{x^2 + 4}$
Common mistake
Putting a constant over a quadratic factor, like $\frac{B}{x^2 + 4}$. With too few unknowns the equations usually have no solution, and you will find coefficients that refuse to match. An irreducible quadratic always gets $Bx + C$. Check any decomposition by adding the pieces back together.
If the numerator's degree is not smaller than the denominator's, divide first (long division), then decompose the remainder.