Algebra & Calculus Integration Tools

Partial Fraction Decomposition Calculator

Decompose proper and improper rational expressions \(\frac{P(x)}{Q(x)}\) step-by-step with polynomial long division, Heaviside cover-up method, and calculus antiderivative integration.

Quick Examples:
Enter in factored form (recommended) or standard polynomial expression.
Proper Rational Fraction (deg P < deg Q)
Partial Fraction Decomposition Result
\[\frac{2}{x - 2} + \frac{3}{x + 1}\]
Calculus Antiderivative / Indefinite Integral
\[\int \frac{P(x)}{Q(x)} dx = 2\ln|x - 2| + 3\ln|x + 1| + C\]
Step-by-Step Mathematical Derivation

What is Partial Fraction Decomposition?

Partial fraction decomposition is an essential algebraic technique that reverses the process of finding a common denominator. When adding fractions like \(\frac{2}{x - 2} + \frac{3}{x + 1}\), you multiply denominators to produce the combined rational function \(\frac{5x - 4}{(x - 2)(x + 1)}\). Partial fraction decomposition takes that single complex fraction and systematically splits it back into its constituent building blocks.

In advanced algebra, calculus, and engineering, decomposing rational expressions is not just an academic exercise—it is a mandatory prerequisite for evaluating difficult integrals, determining inverse Laplace transforms in circuit theory, and solving linear differential equations with constant coefficients.

How to Use the Partial Fraction Decomposition Calculator

Decompose any proper or improper rational expression into simpler fractions in three simple steps:

1

Enter Numerator \(P(x)\)

Input the numerator polynomial (such as 5x - 4 or 2x^3 - x + 1). If degree \(P(x) \ge Q(x)\), the calculator automatically executes polynomial long division first.

2

Enter Denominator \(Q(x)\)

Enter the denominator in expanded or factored format (e.g. x^2 - x - 2, (x - 2)(x + 1)). Quick presets are provided for distinct roots, repeated factors, and improper fractions.

3

View Decomposition & Integral

Instant calculation generates the decomposed partial fractions, undetermined coefficient values (\(A, B, C\)), complete step-by-step derivation, and the exact calculus antiderivative.

Problems This Partial Fraction Calculator Solves

Eliminating Systems of Equations Mistakes

Setting up undetermined coefficients for 3 or 4 unknowns creates tedious linear systems where a single arithmetic error invalidates all constants. Our solver solves the system instantly and shows Heaviside shortcut evaluations.

Auto-Handling Improper Fractions

Students frequently forget that partial fraction templates only work on proper fractions (\(\deg P < \deg Q\)). When \(\deg P \ge \deg Q\), our tool performs polynomial long division first to extract polynomial quotients cleanly.

Instant Calculus Integration Support

Evaluating \(\int \frac{P(x)}{Q(x)} dx\) in Calculus II requires decomposed fractions. The calculator outputs the exact symbolic antiderivative with logarithmic \(\ln|x - c|\) terms.

Control Systems & Laplace Transforms

Electrical engineers decomposing s-domain transfer functions \(H(s)\) can verify inverse Laplace transform poles and residues in seconds.

Key Features & Capabilities

Antiderivative Display

Generates the calculus integral \(\int \frac{P(x)}{Q(x)}dx\) with logarithmic terms.

Heaviside Cover-Up

Demonstrates Heaviside root-plugging evaluation alongside coefficient equating.

Long Division Pre-Pass

Automatically factors out polynomial quotients for improper rational expressions.

Instant Reactive UI

Recalculates on every keystroke with zero calculate button delay.

The Four Canonical Denominator Cases

1. Distinct Linear Factors

When \(Q(x)\) factors into unique terms \((x - r_1)(x - r_2)\dots\), each factor receives an individual constant numerator:

\(\frac{A}{x - r_1} + \frac{B}{x - r_2}\)

2. Repeated Linear Factors

For a factor \((x - r)^k\) of multiplicity \(k\), assign \(k\) terms with ascending denominator powers:

\(\frac{A_1}{x - r} + \frac{A_2}{(x - r)^2} + \dots + \frac{A_k}{(x - r)^k}\)

3. Irreducible Quadratic Factors

A quadratic \((ax^2 + bx + c)\) with negative discriminant (\(b^2 - 4ac < 0\)) cannot be factored over real numbers. It requires a linear numerator:

\(\frac{Ax + B}{ax^2 + bx + c}\)

4. Repeated Irreducible Quadratics

For repeated quadratics \((ax^2 + bx + c)^k\), each power up to \(k\) receives an independent linear numerator:

\(\frac{A_1 x + B_1}{ax^2 + bx + c} + \frac{A_2 x + B_2}{(ax^2 + bx + c)^2}\)

Proper vs. Improper Rational Functions

Before applying any decomposition template, compare the highest powers of the numerator and denominator:

  • Proper Rational Function (\(\deg P < \deg Q\)): The degree of the numerator is strictly smaller than the denominator. You can set up the partial fraction template immediately.
  • Improper Rational Function (\(\deg P \ge \deg Q\)): You must first perform polynomial long division. This converts the fraction into a polynomial quotient \(S(x)\) plus a strictly proper remainder fraction \(\frac{R(x)}{Q(x)}\).

Worked Step-by-Step Example

Problem: Decompose \(\frac{5x - 4}{(x - 2)(x + 1)}\)

Step 1: Denominator has two distinct linear factors: \((x - 2)\) and \((x + 1)\). Set up template:

\[\frac{5x - 4}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}\]

Step 2 (Heaviside Cover-Up for A): Multiply by \((x - 2)\) and evaluate at root \(x = 2\):

\[A = \left.\frac{5x - 4}{x + 1}\right|_{x=2} = \frac{5(2) - 4}{2 + 1} = \frac{6}{3} = 2\]

Step 3 (Heaviside Cover-Up for B): Multiply by \((x + 1)\) and evaluate at root \(x = -1\):

\[B = \left.\frac{5x - 4}{x - 2}\right|_{x=-1} = \frac{5(-1) - 4}{-1 - 2} = \frac{-9}{-3} = 3\]

Step 4: Final decomposition:

\[\frac{5x - 4}{(x - 2)(x + 1)} = \frac{2}{x - 2} + \frac{3}{x + 1}\]

Common Student Mistakes & Pitfalls

1. Forgetting Polynomial Long Division on Improper Fractions Applying partial fractions directly to an improper fraction (e.g. \(\frac{x^2}{x^2 - 1}\)) without first dividing produces completely incorrect coefficients.
2. Omitting Intermediate Powers of Repeated Factors For \((x - 1)^3\), writing only \(\frac{A}{(x - 1)^3}\) is fatal. You must include all powers: \(\frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3}\).
3. Using a Constant Numerator for Irreducible Quadratics An irreducible quadratic like \((x^2 + 4)\) requires a linear numerator \((Ax + B)\), not just a scalar \(A\).

Frequently Asked Questions

What is partial fraction decomposition?
Partial fraction decomposition is an algebraic operation that reverses the process of adding fractions with algebraic denominators. It breaks down a complex rational fraction P(x)/Q(x) into a sum of simpler fractions whose denominators are linear or irreducible quadratic factors of Q(x). This technique is essential for evaluating complex calculus integrals, inverse Laplace transforms, and differential equations.
What should I do if the numerator degree is greater than or equal to the denominator?
When deg(P) >= deg(Q), the fraction is improper. You must first perform polynomial long division to write the fraction as S(x) + R(x)/Q(x), where S(x) is a polynomial quotient and R(x) is a remainder polynomial with deg(R) < deg(Q). Then, apply partial fraction decomposition strictly to the remainder fraction R(x)/Q(x).
What is the Heaviside cover-up method?
The Heaviside cover-up method is a rapid shortcut for finding undetermined coefficients of distinct linear factors. For a factor (x - r), you 'cover up' (remove) (x - r) from the denominator of the original rational expression and evaluate the remaining expression at x = r. The resulting numerical value is the exact numerator coefficient for that partial fraction.
How do you handle repeated linear factors like (x - 2)^3?
For a repeated linear factor (x - r)^k of multiplicity k, the decomposition must include k individual terms with ascending powers in the denominator: A_1/(x - r) + A_2/(x - r)^2 + ... + A_k/(x - r)^k. The highest power coefficient A_k can be found via the cover-up method, while lower powers are solved by equating coefficients or taking derivatives.
What template is used for irreducible quadratic factors like (x^2 + 4)?
An irreducible quadratic factor (ax^2 + bx + c) with b^2 - 4ac < 0 cannot be factored further over the real numbers. Its partial fraction template requires a linear numerator rather than a single constant: (Ax + B)/(ax^2 + bx + c). If the quadratic factor is repeated, e.g. (ax^2 + bx + c)^2, each power receives a linear numerator.