Partial Fraction Decomposition Calculator

Break rational functions into simpler partial fractions for integration

Transform complex rational functions into simpler partial fractions for easier integration, analysis, and problem solving.

Updated June 2026 · How this works

Example calculation — edit any field to use your own numbers

Worth knowing
How It Works
The formula, explained simply

Imagine trying to break apart a complex machine into its basic components. Partial fraction decomposition works the same way with rational functions - it takes one complicated fraction and splits it into several simpler fractions that add up to the original.

The process relies on the fundamental principle that every polynomial can be factored into linear factors (like x-3) and irreducible quadratic factors (like x²+x+1). Once you factor the denominator, you set up a template with unknown coefficients: for each linear factor (x-a), you get a term A/(x-a), and for each quadratic factor, you get (Bx+C)/(quadratic factor).

Solving for these unknown coefficients creates a system of equations. You can either substitute convenient x-values to eliminate terms, or multiply both sides by the original denominator and compare coefficients of like powers. The result transforms complex integration and analysis problems into manageable pieces that follow standard formulas.

When To Use This
Right tool, right situation

Use partial fraction decomposition when integrating rational functions in calculus, especially when the denominator factors nicely into linear or simple quadratic terms. It transforms complex integrals into combinations of logarithmic and arctangent functions with known antiderivatives.

The technique proves essential in differential equations, particularly when applying Laplace transforms. Complex transfer functions in engineering decompose into simpler terms that correspond to standard inverse transforms, making system analysis tractable.

Avoid this method when the denominator factors include high-degree irreducible polynomials or when numerical integration would be more efficient. If the denominator contains factors like x⁴+x²+1 that resist factorization, other integration techniques like trigonometric substitution might prove more practical than forcing partial fraction forms.

Common Mistakes
Why results sometimes look wrong

The most common error is attempting to decompose improper fractions directly without polynomial division. Students see 2x³/(x²-1) and try to write A/(x-1) + B/(x+1), missing that they need to divide first to get 2x + 2x/(x²-1), then decompose only the proper fraction part.

Another frequent mistake involves incorrect partial fraction templates. When the denominator contains (x-2)³, many students write only A/(x-2), forgetting the A₁/(x-2) + A₂/(x-2)² + A₃/(x-2)³ structure. Each power up to the highest requires its own term.

Coefficient solving errors plague even careful students. Setting up the equation A(x+1) + B(x-1) = 3x+5 correctly, then making algebra mistakes when expanding or collecting like terms. Using strategic x-values like x=1 or x=-1 to eliminate unknowns sidesteps much of this arithmetic, but students often expand everything unnecessarily.

The Math
Worked examples and deeper derivation

The mathematical foundation rests on polynomial algebra and the fact that the space of rational functions forms a field. When you have P(x)/Q(x) where P and Q are polynomials with degree of P less than degree of Q, unique partial fraction decomposition always exists.

For linear factors, if (x-r) appears n times in the factorization, you need n terms: A₁/(x-r) + A₂/(x-r)² + ... + Aₙ/(x-r)ⁿ. The coefficients are found by clearing denominators and comparing coefficients, or by using the cover-up method for simple poles.

Irreducible quadratic factors follow a similar pattern but require linear numerators. The general form becomes (Ax+B)/(x²+px+q) for each irreducible quadratic. The proof of uniqueness comes from linear algebra - the coefficient system has a unique solution when the original fraction is proper and the denominator factorization is complete.

Integrating a rational function in calculus
Numerator: 5x + 2, Denominator: x² - 1
Decomposes to 3.5/(x-1) + 1.5/(x+1), making integration straightforward using logarithmic rules instead of complex substitution methods.
Solving differential equations with Laplace transforms
Numerator: 3s + 7, Denominator: s² - 4s + 3
Breaks down complex transfer functions into simpler terms that correspond to known inverse Laplace transforms, enabling solution of the original differential equation.
Simplifying complex algebraic expressions
Numerator: 2x + 8, Denominator: x² + 4x + 3
Converts to 6/(x+1) - 4/(x+3), revealing the behavior near each pole and making limit calculations much clearer.
Expert Unlock
The thing most explanations skip

Professional mathematicians exploit partial fractions beyond integration - they reveal pole locations and residues critical for complex analysis. In engineering, partial fraction decomposition of transfer functions immediately shows system stability through pole placement in the complex plane.

How do I factor the denominator for partial fractions?

What if my fraction is improper?
Perform polynomial long division first to separate the polynomial part from the proper fraction remainder. Only proper fractions (numerator degree less than denominator degree) can be decomposed into partial fractions.
How do I handle repeated linear factors?
For a factor like (x-a)² in the denominator, use both A/(x-a) and B/(x-a)² terms in your decomposition. Each power up to the highest requires its own partial fraction term with unknown coefficients.
What about irreducible quadratic factors?
Quadratic factors that cannot be factored over real numbers require terms like (Ax+B)/(x²+bx+c). The numerator becomes a linear expression rather than just a constant coefficient.

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