Partial fractions undo the addition of fractions. They make rational functions easy to integrate, to take inverse Laplace transforms of and to sum as series. Each linear factor of the denominator contributes one simple fraction.
How to use the Partial Fraction Decomposition Calculator
- Denominator
- Numerator coefficient p (of x² for 3 factors, else of x)
- Numerator coefficient q
- Numerator constant r (3 factors only)
- a
- b
- c
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How it works
For distinct linear factors the coefficient of 1/(x − a) is the rest of the fraction evaluated at x = a (Heaviside’s cover-up rule). For a repeated factor (x − a)², the numerator is rewritten as p(x − a) + (pa + q).
Formula
(px + q)/((x − a)(x − b)) = A/(x − a) + B/(x − b)
A = (pa + q)/(a − b), B = (pb + q)/(b − a)
Worked example
Inputs — Denominator: (x − a)(x − b); Numerator coefficient p (of x² for 3 factors, else of x): 3; Numerator coefficient q: 1; a: 1; b: -2.
Results — Partial fractions = (4/3)/(x − 1) + (5/3)/(x + 2); Steps = (3x + 1) / ((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2)
Cover-up: A = (3·1 + 1) / (1 − (−2)) = (4/3)
B = (3·(−2) + 1) / (−2 − 1) = (5/3).
Key terms
- Rational function
- A ratio of two polynomials.
- Cover-up rule
- Heaviside's shortcut for coefficients of distinct linear factors.
Frequently asked questions
When can I use partial fractions?
When the degree of the numerator is less than the degree of the denominator. Otherwise divide first.
What is the cover-up method?
To find A for (x − a), cover that factor and evaluate the rest of the fraction at x = a.
How do repeated factors work?
(x − a)² needs two terms: A/(x − a) + B/(x − a)².
What about irreducible quadratics?
They need a numerator of the form Bx + C; this calculator covers linear factors.