2.8 Partial Fraction Decomposition

What is a partial fraction?

A partial fraction is a rational expression, that is, it is a ratio of two polynomials, but a very specific type of ratio of two polynomials. Every partial fraction can be written in one of the following two forms, where \(a,b,A,B\) and \(c\) are numbers (\(a\ne0)\) and where \(n\) is a positive integer.

  1. \(\dfrac{A}{(bx+c)^n}\)
  2. \(\dfrac{Ax+B}{(ax^2+bx+c)^n}\)

The reason partial fractions are important in calculus is two-fold.

First, every proper rational expression can be written as the sum of one or more partial fractions. A rational expression is proper if the degree of its numerator is less than the degree of its denominator. When we re-write a proper rational expression as a sum of partial fractions we say that we are "decomposing" the expression into partial fractions, or that we are finding the partial fraction decomposition.

Secondly, certain tasks in calculus such as finding either the anti-derivative or the inverse Laplace transform of a rational function require that it first be decomposed into the sum of partial fractions. This is because there exist rules for finding the anti-derivative and for finding the inverse Laplace transform of partial fractions.

What about improper rational expressions?

If the numerator is of the same or higher degree than the denominator, then we perform long division on the expression and re-write the expression as the sum of a polynomial and a proper rational expression.

For example, \(\dfrac{x^3+x}{x+2}=x^2-2x+5-\dfrac{10}{x+2}\).

The process begins with factoring the denominator

In principle, every polynomial can be factored into polynomial factors of degree one or two--factors of the form \(bx+c\) or of the form \(ax^2+bx+c\). Some second degree or quadratic factors \(ax^2+bx+c\) can be factored into linear factors so we always do that when possible. Quadratic factors which cannot be factored further are said to be irreducible.

When the denominator of a rational expression has been completely factored into linear and quadratic factors, the process of decomposition into partial fractions can begin.

Example 2.7.a

Find the partial fraction decomposition of \( \dfrac{5x+1}{(x-1)(x+2)} \).

The next step is find the appropriate form of the partial fraction decomposition. If all the factors are of the form \(( bx + c )\), then each of the partial fractions will be of the form \( \dfrac{A}{bx+c} \).

\( \dfrac{5x+1}{(x-1)(x+2)} = \dfrac{A}{x-1} +\dfrac{B}{x+2}\)

So this is the form the decomposition must take, but it remains to find the correct numerical values for \(A\) and \(B\).

For terms of the form \(\dfrac{A}{bx+c}\)the constant may be found using Heaviside's Method which we will illustrate here.

To find the value of \(A\) in the partial fraction \( \dfrac{A}{x-1} \) we first multiply the factor \((x-1)\) by \(\dfrac{5x+1}{(x-1)(x+2)}\) to obtain the expression \(\dfrac{5x+1}{x+2}\).

Next we find the zero of the factor \(x-1\) and substitute it into the expression just obtained. The result will be the value of \(A\).

\(A=\dfrac{5x+1}{x+2}\Huge|_{\normalsize x=1}\normalsize =\dfrac{5(1)+1}{1+2}=2\)

To find \(B\) we repeat the process.

\(\dfrac{5x+1}{(x-1)(x+2)}\cdot(x+2)=\dfrac{5x+1}{x-1}\) therefore

\(B=\dfrac{5x+1}{x-1}\Huge|_{\normalsize x=-2}\normalsize =\dfrac{5(-2)+1}{-2-1}=3\)

So the partial fraction decomposition of \( \dfrac{5x+1}{(x-1)(x+2)} = \dfrac{2}{x-1}+\dfrac{3}{x+2}\).

Exercise 2.8.1

Find the partial fraction decomposition of \( \dfrac{x^2+x+3}{x^2+x-6} \)

See Solution

Linear factors \((bx+c)^n\) for \(n > 1\)

When the rational expression contains a factor of the form \(( bx+c )^n\) with \(n > ­1\), the procedure is different. For example:

\[ \dfrac{2x-1}{(x-1)^2}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2} \]

There must be a partial fraction present for each power of \(( bx+c)\) less than or equal to \(n\). Furthermore, only the constant corresponding to \(( bx+c )^n\) may be found by Heaviside's method.

\( \dfrac{2x-1}{(x-1)^2}=2x-1\) so \(B=2x-1\Huge|_{\normalsize x=1}\normalsize =2(1)-1=1\)


So \(\dfrac{2x-1}{(x-1)^2}=\dfrac{A}{x-1}+\dfrac{1}{(x-1)^2}\).

To find the other constant \(A\) multiply both sides by the least common denominator in this equation to clear the fractions:





This last equation is an identity which means it must be true for all values of \(x\). The only way this can be true is if \(A=2\).

So \(\dfrac{2x-1}{(x-1)^2}=\dfrac{2}{x-1}+\dfrac{1}{(x-1)^2}\)

Exercise 2.8.2

Find the partial fraction decomposition of \(\dfrac{x-4}{(x-2)^2}\).

See Solution

Another example

Now let's decompose

\[ \dfrac{-x^2+4x+10}{(x+2)(x+1)^2}=\dfrac{A}{x+2}+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^2} \]

Using Heaviside's method to find \(A\) and \(C\),

\[ A=\dfrac{-x^2+4x+10}{(x+1)^2}\Huge|_{\normalsize x=-2}\normalsize=-2 \] \[ C=\dfrac{-x^2+4x+10}{x+2}\Huge|_{\normalsize x=-1}\normalsize=5 \]


\[ \dfrac{-x^2+4x+10}{(x+2)(x+1)^2}=\dfrac{-2}{x+2}+\dfrac{B}{x+1}+\dfrac{5}{(x+1)^2} \]

Clearing the fractions yields

\[ -x^2+4x+10 = -2(x+1)^2+B(x+2)(x+1)+5(x+2) \]

Since we only need to solve for one number, \(B\), rather than simplify the expression on the right, pick some value of \(x\) (not \(- 1\) or \(- 2\), lest B vanish!) and solve for B. Remember, this equation is an identity, which means that it is true for every value of \(x\). So we pick a value of \(x\) which will make the computation as simple as possible, such as \( x = 0\). Substituting \(x=0\)$ into the equation yields Then the equation reduces to \(10 = - 2 + 2 B + 10\). Thus, \(B = 1\). If you are sceptical that this will always give the correct value of \(B\), then instead of substituting \(x=0\) pick some other value for \(x\). You will still arrive at the value \(B=1\).

So the decomposition is

\[ \dfrac{-x^2+4x+10}{(x+2)(x+1)^2}=\dfrac{-2}{x+2}+\dfrac{1}{x+1}+\dfrac{5}{(x+1)^2} \]

Exercise 2.8.3

Decompose \(\dfrac{5x^2-15x+7}{(x+1)(x-2)^2}\).

See Solution

Irreducible quadratic factors

What about factors in the denominators of the form \((ax^2+bx+c)^n\)? Unfortunately, we cannot use Heaviside's method to find the constants associated with quadratic terms. We will illustrate the process with the following example where \(n=1\).

Decompose \(\dfrac{7x^2-10x+19}{(x-3)(x^2+4)}\).


We find \(A=4\) easily using Heaviside's method but we must find \(B\) and \(C\) by clearing the fractions and simplifying the right side of the equation by collecting like terms:

\(\begin{eqnarray*} \dfrac{7x^2-10x+19}{(x-3)(x^2+4)}&=&\dfrac{4}{x-3}+\dfrac{Bx+C}{x^2+4}\\ 7x^2-10x+19&=&4(x^2+4)+(Bx+C)(x-3)\\ &=&(4+B)x^2+(C-3B)x+(16-3C) \end{eqnarray*}\)

Thus, \(7 = 4 + B\), so \(B = 3\). And since \(C - 3B = -10\), then \(C = -1\). Furthermore, these values are confirmed by \(16-3B=19\).

So the decomposition is \(\dfrac{7x^2-10x+19}{(x-3)(x^2+4)}=\dfrac{4}{x-3}+\dfrac{3x-1}{x^2+4}\).

Exercise 2.8.4

Decompose \(\dfrac{3x^2+7x+10}{(x+2)(x^2+x+2)}\).

See Solution

A quadratic factor with \(n > 1\).

Decompose \(\dfrac{x^3+2x^2+3x+1}{(x^2+1)^2}=\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{(x^2+1)^2}\).

Clearing the fractions yields \(x^3+2x^2+3x+1=(Ax+B)(x^2+1)+Cx+D=Ax^3+Bx^2+(A+C)x+(B+D)\).

So \(A = 1\) and \(B = 2\). Since \(A + C = 3\), \(C = 2\). And since \(B + D = 1, D = -1\).

Thus the partial fraction decomposition is

\[ \dfrac{x^3+2x^2+3x+1}{(x^2+1)^2}=\dfrac{x+2}{x^2+1}+\dfrac{2x-1}{(x^2+1)^2} \]

Exercise 2.8.5

Decompose into partial fractions: \( \dfrac{2x^3-4x^2+11x-1}{(x^2-2x+5)^2} \)

See Solution