Partial Fraction Decomposition

Breaks down rational expressions into simpler, manageable fractions.

Enter the numerator and denominator polynomials below to get the partial fraction expansion. The degree of the numerator must be less than the degree of the denominator.

Examples

Click on an example to load it into the calculator.

Distinct Linear Factors

distinct-linear

A simple case where the denominator factors into unique linear terms.

numerator: [5x - 4]

denominator: [x^2 - x - 2]

Distinct Linear Factors (Cubic)

distinct-linear-cubic

A cubic denominator that factors into three unique linear terms.

numerator: [x^2 + 12x + 12]

denominator: [x^3 - 4x]

Numerator with Constant

improper-requires-division

A case with a constant in the numerator.

numerator: [1]

denominator: [x^2 + x]

Higher Degree Numerator

complex-case

A more complex case with a quadratic numerator.

numerator: [8x^2 - 3x + 10]

denominator: [x^3 - 2x^2 + 4x - 8]

Other Titles
Understanding Partial Fraction Decomposition: A Comprehensive Guide
Master the art of breaking down complex rational functions into simpler components for easier integration and analysis.

What is Partial Fraction Decomposition?

  • Breaking down complex rational expressions.
  • A crucial technique in calculus and engineering.
  • Simplifying fractions for analysis.
Partial fraction decomposition is a procedure in algebra used to express a rational function (a fraction of two polynomials) as a sum of simpler fractions. This technique is invaluable because these simpler fractions are often easier to work with, especially when performing integration or inverse Laplace transforms.
For the decomposition to be possible, the degree of the numerator polynomial must be less than the degree of the denominator polynomial. If it is not, one must first perform polynomial long division.
The Form of Partial Fractions
The form of the partial fraction decomposition depends entirely on the factors of the denominator. The main cases are:
• Distinct Linear Factors: For each factor (ax+b), the decomposition contains a term A/(ax+b).
• Repeated Linear Factors: For each factor (ax+b)^k, the decomposition contains terms A₁/(ax+b) + A₂/(ax+b)² + ... + Aₖ/(ax+b)ᵏ.
• Irreducible Quadratic Factors: For each factor (ax²+bx+c), the decomposition contains a term (Ax+B)/(ax²+bx+c).

Common Decomposition Forms

  • 1 / (x-2)(x+1) -> A/(x-2) + B/(x+1)
  • x / (x-3)^2 -> A/(x-3) + B/(x-3)^2

Step-by-Step Guide to Using the Calculator

  • Inputting your polynomials correctly.
  • Interpreting the calculated results.
  • Using examples to get started quickly.
Our calculator streamlines the decomposition process. Follow these simple steps to get your result.
Inputting Polynomials
Enter the numerator and denominator polynomials into their respective fields. Use 'x' as the variable. Exponents are denoted with the caret symbol '^'. For example, '2x^2 + 3x - 5'.
Calculation and Results
After entering the polynomials, click the 'Calculate' button. The tool will display the decomposed partial fractions. If there are any issues with your input, such as the numerator's degree being too high, an error message will guide you.

Usage Examples

  • Numerator: 'x', Denominator: 'x^2 - 1'
  • Result: 0.5/(x-1) + 0.5/(x+1)

Real-World Applications

  • Solving complex integrals in calculus.
  • Analyzing control systems in engineering.
  • Modeling in physics and chemistry.
Partial fraction decomposition is not just an academic exercise; it has numerous practical applications.
Calculus and Integration
The most common application is in calculus, to find the integral of a rational function. By breaking it into simpler parts, each part can be integrated using basic rules, such as the logarithm rule or power rule.
Engineering and Control Theory
In control systems engineering, transfer functions are often rational functions in the complex variable 's'. Decomposing these functions is a key step in analyzing system behavior and designing controllers. It is used extensively with the inverse Laplace transform to find the time-domain response of a system.

Application Examples

  • Integral of (5x-4)/(x^2-x-2) dx
  • Finding the inverse Laplace transform of F(s) = (s+1)/(s^2+3s+2)

Common Misconceptions and Correct Methods

  • Handling improper rational functions.
  • Correctly forming the decomposition.
  • Solving for the unknown coefficients.
There are several common pitfalls when performing partial fraction decomposition manually.
Improper Fractions
A frequent mistake is attempting to decompose an improper rational function (where the numerator degree is greater than or equal to the denominator degree) directly. The correct first step is always to perform polynomial long division.
Solving for Coefficients
Once the form is set up, finding the coefficients (A, B, C, etc.) can be done by either substituting strategic values for x (like the roots of the denominator) or by expanding the expression and equating the coefficients of like powers of x. The former is often faster and is known as the Heaviside cover-up method for distinct linear roots.

Correction Examples

  • For (x^3)/(x^2-1), first divide to get x + x/(x^2-1), then decompose the remainder.
  • To solve for A in A/(x-r), multiply by (x-r) and then substitute x=r.

Mathematical Derivation and Theory

  • The theory behind the decomposition.
  • Proof of existence and uniqueness.
  • The Heaviside cover-up method explained.
The possibility of decomposing a rational function is guaranteed by a theorem in algebra which states that any proper rational function can be written as a unique sum of partial fractions.
The Heaviside Cover-Up Method
For a proper rational function P(x)/Q(x) where Q(x) has distinct linear factors (x - r₁)(x - r₂)...(x - rₙ), the decomposition is A₁/(x-r₁) + ... + Aₙ/(x-rₙ). The coefficient Aₖ can be found by 'covering up' the factor (x - rₖ) in the original fraction and evaluating the rest at x = rₖ.
Mathematically, Aₖ = P(rₖ) / Q'(rₖ), where Q'(x) is the derivative of Q(x). This provides a quick way to find coefficients without solving a large system of linear equations.

Method Example

  • To find A for 1/((x-2)(x+1)) = A/(x-2) + B/(x+1), calculate A = 1/(2+1) = 1/3.