The **partial fraction decomposition** or also commonly known as **partial fraction expansion** is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

The importance of the partial fraction decomposition lies in the fact that it provides an algorithm for computing the __antiderivative of a rational function__.

In symbols, one can use *partial fraction expansion* to change a rational fraction in the form

where *ƒ* and *g* are polynomials, into an expression of the form

where *g*_{j} (*x*) are polynomials that are factors of *g*(*x*), and are in general of lower degree.
Thus, the partial fraction decomposition may be seen as the inverse procedure of the more elementary operation of addition of rational fractions, which produces a single rational fraction with a numerator and denominator usually of high degree. Thus, the outcome of a full partial fraction expansion expresses that fraction as a sum of fractions, where:

- the denominator of each term is a power of an irreducible (not factorable) polynomial and
- the numerator is a polynomial of smaller degree than that irreducible polynomial.

## Basic principles Edit

The basic principles involved are quite simple,

Assume a rational function *R*(*x*) = *ƒ*(*x*)/*g*(*x*) in one indeterminate *x* has a denominator that factors as

over a field *K* (we can take this to be real numbers, or complex numbers). If *P* and *Q* have no common factor, then *R* may be written as

## Procedure Edit

Given two polynomials and , where the *α*_{i} are distinct constants and deg *P* < *n*, partial fractions are generally obtained by supposing that

and solving for the *c*_{i} constants, by substitution, by equating the coefficients of terms involving the powers of *x*, or otherwise. (This is a variant of the method of undetermined coefficients.)

A more direct computation, consists in writing

where is the derivative of the polynomial .

## Examples Edit

### Example 1 Edit

Here, the denominator splits into two distinct linear factors:

so we have the partial fraction decomposition

Multiplying through by *x*^{2} + 2*x* − 3, we have the polynomial identity

Substituting *x* = −3 into this equation gives *A* = −1/4, and substituting *x* = 1 gives *B* = 1/4, so that

### Example 2 Edit

After performing long-division, we have

Since (−4)^{2} − 4(8) = −16 < 0, *x*^{2} − 4*x* + 8 is irreducible, and so

Multiplying through by *x*^{3} − 4*x*^{2} + 8*x*, we have the polynomial identity

Taking *x* = 0, we see that 16 = 8*A*, so *A* = 2. Comparing the *x*^{2} coefficients, we see that 4 = *A* + *B* = 2 + *B*, so *B* = 2. Comparing linear coefficients, we see that −8 = −4*A* + *C* = −8 + *C*, so *C* = 0. Altogether,

### Example 3 Edit

After performing long-division and factoring the denominator, we have

The partial fraction decomposition takes the form

Multiplying through by (*x* − 1)^{3}(*x*^{2} + 1)^{2} we have the polynomial identity

Taking *x* = 1 gives 4 = 4*C*, so *C* = 1. Similarly, taking *x* = *i* gives 2 + 2*i* = (*Fi* + *G*)(2 + 2*i*), so *Fi* + *G* = 1, so *F* = 0 and *G* = 1 by equating real and imaginary parts. With *C* = *G* = 1 and *F* = 0, taking *x* = 0 we get *A* − *B* + 1 − *E* − 1 = 0, thus *E* = *A* − *B*.

We now have the identity

Expanding and sorting by exponents of x we get

We can now compare the coefficients and see that

with *A* = 2 − *D* and −*A* −3 *D* =−4 we get *A* = *D* = 1 and so *B* = 0, furthermore is *C* = 1, *E* = *A* − *B* = 1, *F* = 0 and *G* = 1.

The partial fraction decomposition of *ƒ*(*x*) is thus

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at *x=1* and at *x=i* in the above polynomial identity. (To this end, recall that the derivative at *x=a* of *(x−a) ^{m}p(x)* vanishes if

*m > 1*and it is just

*p(a)*if

*m=1*.) Thus, for instance the first derivative at

*x=1*gives

that is *8 = 2B + 8* so *B=0*.