Last updated October 4, 2026 · Report an error
Partial fraction decomposition breaks a complicated rational expression like (5x+3)/((x-1)(x-2)) into a sum of simpler fractions — a key technique before integrating rational functions in calculus. This tool decomposes it using the fast "cover-up" method.
How it works
For a denominator of distinct linear factors, each numerator coefficient is found by covering up its corresponding factor and substituting that factor's root into everything else — no system of equations to solve by hand.
- Enter denominator degree (number of linear factors).
- Enter r₁ (first root).
- Enter r₂ (second root).
- Enter r₃ (third root, only if degree 3).
- Enter numerator: x² coefficient (only if degree 3).
- Enter numerator: x coefficient.
- Enter numerator: constant.
- Click Calculate to see your results.
Examples
A 2-factor example
(5x+3)/((x-1)(x-2)) decomposes to -8/(x-1) + 13/(x-2).
Cross-checking the answer
Recombining -8/(x-1)+13/(x-2) over a common denominator reproduces exactly 5x+3 in the numerator, confirming the decomposition.
Who should use it
- Calculus coursework preparing to integrate rational functions.
- Checking partial fraction decomposition homework.
Industry applications
- Calculus and engineering education (Laplace transforms, control systems)
Advantages
- Uses the fast cover-up method instead of solving a full system of equations.
- Shows the substitution for every coefficient explicitly.
Limitations
- Limited to distinct linear factors (up to a degree-3 denominator) — repeated or irreducible quadratic factors aren't supported.
Common mistakes to avoid
- Applying the cover-up method to a denominator with a repeated factor, where it doesn't directly apply without modification.
Best practices
- Always double-check a decomposition by recombining the partial fractions over a common denominator and confirming it matches the original numerator.
Tips
- Once decomposed, each term A/(x-r) integrates directly to A·ln|x-r| — the whole point of doing the decomposition in the first place.
Frequently asked questions
For each linear factor (x-rᵢ), you mentally cover it up in the original denominator and substitute x=rᵢ into what remains — that gives the numerator coefficient for that term directly, without solving a system of equations.
The cover-up method relies on the roots being distinct so each substitution isolates exactly one unknown coefficient — repeated or irreducible quadratic factors require an extended version of the technique.