Last updated: 2026-07-02

Partial fraction decomposition

Partial fraction decomposition rewrites a single rational function as a sum of simpler fractions whose denominators are the factors of the original denominator. For a proper fraction — one whose numerator has lower degree than its denominator — with two distinct linear factors, the decomposition takes the form

$$\frac{p\,x + q}{(x - r_1)(x - r_2)} = \frac{A}{x - r_1} + \frac{B}{x - r_2}.$$

Splitting the fraction this way makes it far easier to integrate, to take an inverse Laplace transform, or to expand as a series.

The cover-up method

The fastest way to find $A$ and $B$ is the cover-up method (also called the Heaviside cover-up).

Step 1 — Find A. Cover the factor $(x - r_1)$ in the denominator and evaluate what is left at $x = r_1$:

$$A = \left.\frac{p\,x + q}{x - r_2}\right|_{x = r_1} = \frac{p\,r_1 + q}{r_1 - r_2}.$$

Step 2 — Find B. Cover the factor $(x - r_2)$ and evaluate the rest at $x = r_2$:

$$B = \left.\frac{p\,x + q}{x - r_1}\right|_{x = r_2} = \frac{p\,r_2 + q}{r_2 - r_1}.$$

Why cover-up works

Multiply both sides of the decomposition by $(x - r_1)$:

$$\frac{p\,x + q}{x - r_2} = A + \frac{B\,(x - r_1)}{x - r_2}.$$

Now set $x = r_1$. The second term contains the factor $(x - r_1)$, so it becomes zero, leaving $A = (p\,r_1 + q)/(r_1 - r_2)$ directly. The same argument with $(x - r_2)$ isolates $B$.

Worked example

Decompose $\dfrac{3x - 1}{(x - 1)(x + 2)}$.

Here $p = 3$, $q = -1$, $r_1 = 1$, $r_2 = -2$.

$$A = \frac{3(1) - 1}{1 - (-2)} = \frac{2}{3}$$ $$B = \frac{3(-2) - 1}{-2 - 1} = \frac{-7}{-3} = \frac{7}{3}$$

So

$$\frac{3x - 1}{(x - 1)(x + 2)} = \frac{2/3}{x - 1} + \frac{7/3}{x + 2}.$$

Check: combining over the common denominator,

$$\frac{2}{3}(x + 2) + \frac{7}{3}(x - 1) = \frac{2x + 4 + 7x - 7}{3} = \frac{9x - 3}{3} = 3x - 1. \checkmark$$

A useful shortcut

For the numerator $p\,x + q$, the two coefficients always satisfy $A + B = p$. Matching the leading term of the recombined numerator forces the sum of $A$ and $B$ to equal the numerator's $x$-coefficient, which is a quick sanity check on your answer.

Distinct roots are required

The form $A/(x - r_1) + B/(x - r_2)$ assumes two different linear factors.

Denominator	Correct form
$(x - r_1)(x - r_2)$, distinct	$\dfrac{A}{x - r_1} + \dfrac{B}{x - r_2}$
$(x - r_1)^2$, repeated	$\dfrac{A}{x - r_1} + \dfrac{B}{(x - r_1)^2}$

The cover-up formula divides by $r_1 - r_2$, which is zero when the roots coincide, so a repeated factor needs the second form and is outside the scope of this calculator.

Use in integration

The main payoff is in calculus. A rational function is awkward to integrate directly, but each partial fraction integrates immediately:

$$\int \frac{A}{x - r}\,dx = A \ln|x - r| + C.$$

So once the fraction is decomposed,

$$\int \frac{p\,x + q}{(x - r_1)(x - r_2)}\,dx = A \ln|x - r_1| + B \ln|x - r_2| + C.$$

Applications

• Integration of rational functions in calculus.
• Inverse Laplace transforms in differential equations and control theory.
• Series expansions and generating functions in combinatorics.
• Signal processing, where transfer functions are split into first-order terms.

Frequently Asked Questions (FAQ)

What is partial fraction decomposition?

Partial fraction decomposition rewrites a single rational function as a sum of simpler fractions whose denominators are the factors of the original denominator. For a proper fraction (px + q) / ((x − r₁)(x − r₂)) with two distinct linear factors, the decomposition is A/(x − r₁) + B/(x − r₂). Splitting the fraction this way makes it far easier to integrate, take inverse Laplace transforms, or expand as a power series.

How does the cover-up method find A and B?

To find A, cover the factor (x − r₁) in the denominator and evaluate what remains at x = r₁. Since the second term A·(x − r₁)/(x − r₁) survives only when its own factor is covered, everything else vanishes at that value. The result is A = (p·r₁ + q) / (r₁ − r₂). By symmetry, B = (p·r₂ + q) / (r₂ − r₁). The method avoids setting up and solving a linear system.

Why must the two roots be distinct?

The form A/(x − r₁) + B/(x − r₂) assumes two different linear factors. If r₁ = r₂, the denominator is a repeated factor (x − r₁)², and the correct decomposition uses the form A/(x − r₁) + B/(x − r₁)². The cover-up formula divides by (r₁ − r₂), which is zero when the roots coincide, so this calculator requires distinct roots.

How is this used in integration?

A rational function is hard to integrate directly, but each partial fraction A/(x − r) integrates immediately to A·ln|x − r| + C. So once (px + q) / ((x − r₁)(x − r₂)) is decomposed into A/(x − r₁) + B/(x − r₂), the integral is A·ln|x − r₁| + B·ln|x − r₂| + C. This is the standard first step for integrating rational functions in calculus.