In mathematics, partial fractions are used in real-variable integral calculus to find real-valued antiderivatives of rational functions. The partial fraction decomposition of real rational functions is also used for Laplace transforms.
This article describes the general method for obtaining the partial fraction decomposition of any real rational function. The proof of the existence of the partial fraction decomposition over an arbitrary field is not given here. For a sketch of the general proof, see partial fraction.
For applications of partial fraction decomposition over R, see
- Partial fractions in integration
- Partial fractions in Laplace transforms
Table of contents |
2 The algorithm 3 Some examples |
Let f(x) be any rational function over the real numbers. In other words, suppose there exist real polynomials p(x) and q(x)≠ 0, such that
General result
By removing the leading coefficient of q(x), we may assume without loss of generality that q(x) is monic. By the fundamental theorem of algebra, we can write
Then the partial fraction decomposition of f(x) is the following.
The most straightforward method is to multiply through by the common denominator q(x). We then obtain an equation of polynomials whose left-hand side is simply p(x) and whose right-hand side has coefficients which are linear expressions of the constants Air, Bir, and Cir. Since two polynomials are equal if and only if their corresponding coefficients are equal, we can equate the coefficients of like terms. In this way, a system of linear equations is obtained which always has a unique solution. This solution can be found using any of the standard methods of linear algebra. However, this is often not the best way to go when computing by pencil and paper, and there are other ways ("tricks") to obtain the constants.
The partial fraction decomposition of a real rational function can be obtained by the following procedure:
Example 1
The algorithm
Some examples
Here, the denominator splits into two distinct linear factors:
so we have the partial fraction decomposition
Multiplying through by x2 + 2x - 3, we have the polynomial identity
Substituting x = -3 and x = 1 into this equation gives A = -1/4 and B = 1/4, so that
Example 2
After long-division, we have
Since (-4)2-4(8) = -16 < 0, x2 - 4x + 8 is irreducible, and so
Multiplying through by x3 - 4x2 + 8x, we have the polynomial identity
Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that -8 = -4A + C = -8 + C, so C = 0. Altogether,
The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.
Example 3