Partial Fractions
Partial fraction decomposition writes a rational expression as a sum of simpler fractions whose denominators are the factors of the original denominator. In brief: (5x + 7)/((x + 1)(x + 2)) = 2/(x + 1) + 3/(x + 2). The reverse operation is adding fractions with a common denominator. The decomposition is used to integrate rational functions, to sum series and to invert transforms.
Adding fractions combines pieces into one, and partial fractions take one fraction apart. For a proper fraction (the degree of the numerator is less than the degree of the denominator), the form of the decomposition depends on the factors of the denominator: distinct linear factors, repeated linear factors or irreducible quadratic factors. This guide describes each case with a method for finding the coefficients, worked examples and a table of the forms.
When can you use partial fractions?
The method needs a rational expression P(x)/Q(x) in which the degree of P is less than the degree of Q (a proper fraction) and the denominator is factored into linear factors and irreducible quadratic factors.
- Factor the denominator completely.
- Write the form of the decomposition with unknown constants.
- Multiply both sides by the denominator.
- Find the constants by substituting convenient values of x or by comparing coefficients.
- Check by adding the fractions or by testing a value.
What if the denominator has distinct linear factors?
Each factor (x − a) gives a term A/(x − a).
Example 1: (5x + 7)/((x + 1)(x + 2))
Write (5x + 7)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2). Multiply by (x + 1)(x + 2): 5x + 7 = A(x + 2) + B(x + 1). Substitute x = −1: 2 = A · 1, so A = 2. Substitute x = −2: −3 = B · (−1), so B = 3. The decomposition is 2/(x + 1) + 3/(x + 2). Check: [2(x + 2) + 3(x + 1)]/((x + 1)(x + 2)) = (5x + 7)/(…) ✓. This substitution trick is the cover-up method: cover the factor and evaluate the rest at the zero.
Example 2: three factors
(x2 + 1)/(x(x − 1)(x + 1)) = A/x + B/(x − 1) + C/(x + 1). At x = 0: 1 = A(−1)(1), so A = −1. At x = 1: 2 = B · 1 · 2, so B = 1. At x = −1: 2 = C · (−1)(−2), so C = 1. So the result is −1/x + 1/(x − 1) + 1/(x + 1).
What if a linear factor is repeated?
A factor (x − a)k gives k terms: A1/(x − a) + A2/(x − a)2 + … + Ak/(x − a)k.
Example 3: (3x + 5)/((x − 1)(x + 2)2)
Form: A/(x − 1) + B/(x + 2) + C/(x + 2)2. Multiply: 3x + 5 = A(x + 2)2 + B(x − 1)(x + 2) + C(x − 1). At x = 1: 8 = 9A, so A = 8/9. At x = −2: −1 = −3C, so C = 1/3. For B compare the coefficients of x2: 0 = A + B, so B = −8/9. Check at x = 0: 5 = 4A − 2B − C = 32/9 + 16/9 − 3/9 = 45/9 = 5 ✓.
What if there is an irreducible quadratic factor?
An irreducible quadratic factor (ax2 + bx + c, with a negative discriminant) gives a term with a linear numerator: (Ax + B)/(ax2 + bx + c).
Example 4: (2x2 + x + 3)/((x − 1)(x2 + 1))
Form: A/(x − 1) + (Bx + C)/(x2 + 1). Multiply: 2x2 + x + 3 = A(x2 + 1) + (Bx + C)(x − 1). At x = 1: 6 = 2A, so A = 3. Expand and compare: A + B = 2 (coefficient of x2), so B = −1. The constant term: A − C = 3, so C = 0. Check the coefficient of x: C − B = 0 + 1 = 1 ✓. The decomposition is 3/(x − 1) − x/(x2 + 1).
What if the fraction is improper?
Divide first to obtain a polynomial plus a proper fraction, then decompose the proper fraction.
Example 5: (x3 + 2x2 + 1)/(x2 − 1)
Divide: x3 + 2x2 + 1 = (x2 − 1)(x + 2) + (x + 3). So the fraction is x + 2 + (x + 3)/((x − 1)(x + 1)). Decompose: (x + 3)/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1). At x = 1: 4 = 2A, so A = 2. At x = −1: 2 = −2B, so B = −1. The result: x + 2 + 2/(x − 1) − 1/(x + 1).
Example 6: an application to a sum
1/(n(n + 1)) = 1/n − 1/(n + 1). Therefore the sum 1/(1 · 2) + 1/(2 · 3) + … + 1/(n(n + 1)) telescopes to 1 − 1/(n + 1) = n/(n + 1). For n = 99: 99/100.
Example 7: integration preview
∫ (5x + 7)/((x + 1)(x + 2)) dx = ∫ [2/(x + 1) + 3/(x + 2)] dx = 2 ln|x + 1| + 3 ln|x + 2| + C.
Table of decomposition forms
| Factor in the denominator | Terms in the decomposition | Example |
|---|---|---|
| x − a | A/(x − a) | 2/(x + 1) |
| (x − a)2 | A/(x − a) + B/(x − a)2 | 1/(x + 2) + 1/(x + 2)2 |
| (x − a)3 | A/(x − a) + B/(x − a)2 + C/(x − a)3 | three terms |
| x2 + px + q (irreducible) | (Ax + B)/(x2 + px + q) | (Ax + B)/(x2 + 1) |
| (x2 + px + q)2 | (Ax + B)/(…) + (Cx + D)/(…)2 | two terms |
Where does the method come from?
The decomposition was introduced as a tool for integration by Johann Bernoulli and Leibniz at the beginning of the eighteenth century, who used it to integrate rational functions. Euler later gave a systematic treatment. The result that every proper rational function over the reals can be written as a sum of the described terms depends on the fact that every real polynomial factors into linear and irreducible quadratic factors, a consequence of the fundamental theorem of algebra.
Where is it used today?
Engineers use partial fractions to find the inverse Laplace transform of a rational function, which gives the response of a system as a sum of exponentials. Probabilists use them to find the distribution of a sum of random variables with rational generating functions. In algebra, the method is used to find closed formulas for sequences defined by recurrences.
Which mistakes are common?
Forgetting the division for an improper fraction
The degree of the numerator must be smaller than the degree of the denominator. Otherwise divide first.
Using a constant numerator for a quadratic factor
An irreducible quadratic needs a linear numerator Ax + B.
Missing terms for repeated factors
(x + 2)2 needs two terms, one with (x + 2) and one with (x + 2)2.
Not checking the result
Substitute a value such as x = 0 into both sides, or add the fractions back together.