Partial Fraction Decomposition Calculator
Our free algebra calculator solves partial fraction decomposition problems. Get worked examples, visual aids, and downloadable results.
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The rational expression is decomposed into a sum of simpler fractions. Constants A and B are found using the cover-up (Heaviside) method: set each factor to zero and evaluate the remaining expression. The resulting fractions are easier to integrate, differentiate, and analyze.
Last reviewed: December 2025
Worked Examples
Example 1: Simple Partial Fraction Decomposition
Example 2: Integration via Partial Fractions
Background & Theory
The Partial Fraction Decomposition Calculator applies the following established principles and formulas. Mathematics rests on a hierarchy of number systems, each extending the previous. The natural numbers (1, 2, 3, ...) support counting and ordering. The integers add negative values and zero, enabling subtraction without restriction. The rational numbers, expressible as p/q where p and q are integers and q is nonzero, close the system under division. The real numbers fill the gaps left by irrationals such as the square root of 2 or pi, forming a complete ordered field. The complex numbers, written as a + bi where i is the square root of negative one, complete the algebraic closure of the reals and allow every polynomial to have a root. Prime factorization states that every integer greater than one is uniquely expressible as a product of primes, a result known as the Fundamental Theorem of Arithmetic. Computing the greatest common divisor (GCD) of two integers relies most efficiently on the Euclidean algorithm: repeatedly replace the larger number with the remainder when it is divided by the smaller, until the remainder is zero. The last nonzero remainder is the GCD. The least common multiple (LCM) follows from the identity LCM(a, b) = |a * b| / GCD(a, b). Modular arithmetic defines equivalence classes of integers that share the same remainder under division by a modulus n. Fermat's Little Theorem and Euler's Theorem arise from this structure and underpin modern cryptography. Logarithms are the inverses of exponential functions. If b raised to the power x equals y, then the logarithm base b of y equals x. The natural logarithm uses base e, approximately 2.71828. Combinatorics counts arrangements and selections. The number of ordered arrangements (permutations) of r objects from n distinct objects is nPr = n! / (n - r)!. The number of unordered selections (combinations) is nCr = n! / (r! * (n - r)!). Pascal's triangle arranges these binomial coefficients so that each entry equals the sum of the two entries directly above it. The Fibonacci sequence, defined by F(1) = 1, F(2) = 1, and F(n) = F(n-1) + F(n-2), appears throughout nature and connects deeply to the golden ratio via Binet's formula.
History
The history behind the Partial Fraction Decomposition Calculator traces back through the following developments. Mathematics as a systematic discipline traces to ancient Mesopotamia. Babylonian clay tablets dating to around 1800 BCE demonstrate knowledge of quadratic equations, Pythagorean triples, and base-60 arithmetic, suggesting a practical mathematical tradition far preceding Greek formalism. Euclid of Alexandria compiled the Elements around 300 BCE, establishing the axiomatic method that would define rigorous mathematics for over two thousand years. His work organized plane geometry, number theory, and proportion into logically chained propositions derived from a small set of postulates. The algorithm bearing his name for computing GCDs appears in Book VII and remains in use today. In the 9th century, the Persian scholar Muhammad ibn Musa Al-Khwarizmi wrote Al-Kitab al-mukhtasar fi hisab al-jabr wal-muqabala, the treatise whose title gave algebra its name. He systematized the solution of linear and quadratic equations and described procedures that operated on unknowns as objects, a conceptual leap away from purely numerical calculation. Rene Descartes introduced coordinate geometry in 1637 by uniting algebra and Euclidean geometry, allowing curves to be studied through equations. This synthesis set the stage for calculus. Isaac Newton and Gottfried Wilhelm Leibniz independently developed calculus during the 1660s and 1670s, triggering a priority dispute that lasted decades and divided British and Continental mathematicians. Carl Friedrich Gauss proved the Fundamental Theorem of Algebra in 1799, showing that every nonconstant polynomial has at least one complex root. His Disquisitiones Arithmeticae of 1801 established modern number theory. David Hilbert's formalist program at the turn of the 20th century sought to place all of mathematics on an explicit axiomatic foundation, a project that Kurt Godel's incompleteness theorems of 1931 showed to be fundamentally limited. Alan Turing's work in the 1930s on computability introduced the theoretical model of the stored-program computer and linked mathematical logic directly to the limits of algorithmic calculation. His proof that no algorithm can decide in general whether an arbitrary program will halt or run forever placed fundamental boundaries on what mathematics can mechanically determine, and it opened the discipline now known as theoretical computer science.
Frequently Asked Questions
Formula
(ax + b) / ((c1x - r1)(c2x - r2)) = A/(c1x - r1) + B/(c2x - r2)
The rational expression is decomposed into a sum of simpler fractions. Constants A and B are found using the cover-up (Heaviside) method: set each factor to zero and evaluate the remaining expression. The resulting fractions are easier to integrate, differentiate, and analyze.
Worked Examples
Example 1: Simple Partial Fraction Decomposition
Problem: Decompose (5x + 3) / ((x - 1)(x + 3)) into partial fractions.
Solution: Form: (5x + 3) / ((x - 1)(x + 3)) = A/(x - 1) + B/(x + 3)\n\nCover-up method:\nFor A: Set x = 1: A = (5(1) + 3) / (1 + 3) = 8/4 = 2\nFor B: Set x = -3: B = (5(-3) + 3) / (-3 - 1) = (-12)/(-4) = 3\n\nResult: 2/(x - 1) + 3/(x + 3)\n\nVerify at x = 0: Original = 3/((-1)(3)) = -1\nDecomposed = 2/(-1) + 3/3 = -2 + 1 = -1. Verified!
Result: (5x + 3)/((x-1)(x+3)) = 2/(x-1) + 3/(x+3)
Example 2: Integration via Partial Fractions
Problem: Find the integral of (7x - 1) / ((2x - 3)(x + 2)) dx.
Solution: Decompose: A/(2x - 3) + B/(x + 2)\nCover up (2x-3), set x = 3/2: A = (7(3/2) - 1)/(3/2 + 2) = (9.5)/(3.5) = 19/7\nCover up (x+2), set x = -2: B = (7(-2) - 1)/(2(-2) - 3) = (-15)/(-7) = 15/7\n\nIntegral = (19/7)(1/2)ln|2x-3| + (15/7)ln|x+2| + C\n= (19/14)ln|2x-3| + (15/7)ln|x+2| + C
Result: Integral = (19/14)ln|2x-3| + (15/7)ln|x+2| + C
Frequently Asked Questions
What is partial fraction decomposition and when is it used?
Partial fraction decomposition is an algebraic technique that breaks a complex rational expression into a sum of simpler fractions. The idea is to reverse the process of adding fractions with unlike denominators. For example, instead of working with (5x + 3)/((x - 1)(x + 3)), we decompose it into A/(x - 1) + B/(x + 3). This technique is essential in calculus for integrating rational functions, because simpler fractions are much easier to integrate using basic logarithm and arctangent rules. It is also heavily used in Laplace transforms for solving differential equations, in signal processing for analyzing transfer functions, and in control theory for analyzing system behavior.
How do you find the constants A and B in partial fraction decomposition?
There are two main methods for finding the constants. The cover-up method (also called Heaviside method) is the fastest for distinct linear factors. Set each factor equal to zero and substitute the resulting x value into the remaining expression. For (5x + 3)/((x-1)(x+3)): cover up (x-1) and set x=1 to get A = (5+3)/(1+3) = 2. Cover up (x+3) and set x=-3 to get B = (-15+3)/(-3-1) = 3. The second method is comparing coefficients: write A(x+3) + B(x-1) = 5x+3, expand, and match the coefficients of x and the constant terms to create a system of equations. Both methods give identical results, but cover-up is faster for simple cases.
What types of denominators require different decomposition forms?
The form of the decomposition depends on the types of factors in the denominator. Distinct linear factors like (x-1)(x+2) yield A/(x-1) + B/(x+2). Repeated linear factors like (x-1)^2 require A/(x-1) + B/(x-1)^2. Irreducible quadratic factors like (x^2+1) require (Ax+B)/(x^2+1) because the numerator must be one degree less than the denominator factor. Repeated irreducible quadratics like (x^2+1)^2 require (Ax+B)/(x^2+1) + (Cx+D)/(x^2+1)^2. The total number of unknown constants always equals the degree of the denominator. Getting the form right is crucial because using the wrong form will lead to an inconsistent system of equations.
How does partial fraction decomposition help with integration?
Partial fraction decomposition transforms difficult integrals into sums of elementary integrals. The integral of A/(x-r) is A*ln|x-r| + C, which is simple. Without decomposition, integrating (5x+3)/((x-1)(x+3)) would require complicated substitutions or other techniques. After decomposing into 2/(x-1) + 3/(x+3), the integral becomes 2*ln|x-1| + 3*ln|x+3| + C, which is straightforward. For irreducible quadratic factors, the decomposed fractions integrate into arctangent functions. This is why partial fractions is taught as a core integration technique in every calculus course, and it is one of the few methods that systematically handles all rational function integrals.
Can partial fractions be applied to improper rational expressions?
An improper rational expression has a numerator with degree greater than or equal to the degree of the denominator. Before applying partial fraction decomposition, you must first perform polynomial long division to extract the polynomial part. For example, (x^3 + 2x + 1)/(x^2 - 1) is improper because the numerator degree (3) exceeds the denominator degree (2). Long division gives x + (3x + 1)/(x^2 - 1). Now you apply partial fraction decomposition only to the proper remainder fraction (3x + 1)/((x-1)(x+1)). This extra step is essential because partial fraction decomposition only works for proper fractions where the numerator degree is strictly less than the denominator degree.
How are partial fractions used in Laplace transforms?
In solving differential equations using Laplace transforms, you often end up with a rational function F(s) in the s-domain that needs to be inverted back to the time domain. Partial fraction decomposition breaks F(s) into simple fractions whose inverse Laplace transforms are known. For example, F(s) = (s + 5)/((s + 1)(s + 3)) decomposes into 2/(s+1) + (-1)/(s+3), which inverts to f(t) = 2e^(-t) - e^(-3t). Without partial fractions, finding the inverse Laplace transform of complex rational expressions would be extremely difficult. This technique is fundamental in electrical engineering for circuit analysis and in control systems for determining system response characteristics.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy