Partial Fraction Decomposition Calculator
Our free algebra calculator solves partial fraction decomposition problems. Get worked examples, visual aids, and downloadable results.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Partial Fraction Decomposition Calculator
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Formula: (ax + b) / ((c1x - r1)(c2x - r2)) = A/(c1x - r1) + B/(c2x - r2)
Worked example — (5x + 3)/((x-1)(x+3)) = 2/(x-1) + 3/(x+3)
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) Cover-up method: For A: Set x = 1: A = (5(1) + 3) / (1 + 3) = 8/4 = 2 For B: Set x = -3: B = (5(-3) + 3) / (-3 - 1) = (-12)/(-4) = 3 Result: 2/(x - 1) + 3/(x + 3) Verify at x = 0: Original = 3/((-1)(3)) = -1 Decomposed = 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) Cover up (2x-3), set x = 3/2: A = (7(3/2) - 1)/(3/2 + 2) = (9.5)/(3.5) = 19/7 Cover up (x+2), set x = -2: B = (7(-2) - 1)/(2(-2) - 3) = (-15)/(-7) = 15/7 Integral = (19/7)(1/2)ln|2x-3| + (15/7)ln|x+2| + C = (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.
What is the cover-up method and why is it so efficient?
The cover-up method, attributed to Oliver Heaviside, provides a remarkably fast way to find partial fraction coefficients for distinct linear factors. The idea is simple: to find the coefficient for the term A/(x-r), mentally cover up the factor (x-r) in the original fraction and evaluate the remaining expression at x=r. For example, in (2x+1)/((x-3)(x+1)), to find A: cover up (x-3) and plug in x=3 to get (7)/(4) = 7/4. To find B: cover up (x+1) and plug in x=-1 to get (-1)/(-4) = 1/4. This works because setting x=r makes (x-r) = 0, which eliminates all terms except the one you want, effectively isolating the coefficient. The method is much faster than setting up and solving systems of equations.
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.
What happens when the denominator has complex roots?
When the denominator has irreducible quadratic factors (those with no real roots), the partial fraction decomposition uses a linear numerator of the form (Ax + B) instead of a constant. For example, (3x + 5)/((x-2)(x^2+4)) decomposes as A/(x-2) + (Bx + C)/(x^2+4). The quadratic factor x^2+4 has complex roots (plus or minus 2i), but we keep the real form because it integrates neatly: the Bx part gives a logarithm via substitution, and the C part gives an arctangent. You could alternatively decompose into complex linear factors, but real-coefficient forms are standard in calculus courses and engineering applications because they produce real-valued results.
How do you handle repeated factors in partial fraction decomposition?
Repeated factors require a term for each power of the repeated factor. For (3x+2)/((x-1)^3), the decomposition is A/(x-1) + B/(x-1)^2 + C/(x-1)^3. You need three terms because the factor (x-1) is raised to the third power. The cover-up method only works directly for the highest power term C: setting x=1 gives C = (3+2)/1 = 5. Finding A and B requires either comparing coefficients or substituting two additional convenient values of x. For integration purposes, the terms with powers of (x-1) in the denominator integrate using the power rule rather than the logarithm rule: the integral of 1/(x-1)^n is -1/((n-1)(x-1)^(n-1)) for n greater than 1.
What is the connection between partial fractions and polynomial identities?
Partial fraction decomposition fundamentally relies on polynomial identities. When you write (ax+b)/((x-r1)(x-r2)) = A/(x-r1) + B/(x-r2), multiplying both sides by the denominator gives the polynomial identity ax+b = A(x-r2) + B(x-r1). This identity must hold for ALL values of x, which means the coefficients of each power of x on both sides must be equal. This is the comparing coefficients method. Alternatively, since the identity holds for all x, we can choose specific convenient values of x (like x = r1 and x = r2) to create simple equations. This dual perspective, viewing the equation as either a universal identity or a set of specific evaluations, gives us the flexibility to use whichever method is computationally simpler for a given problem.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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