Download Partial Fraction Decomposition Examples

Partial Fraction Decomposition Examples
Partial fraction decomposition is a technique used to convert a fraction with a polynomial numerator
and a polynomial denominator into the sum of two or more simpler fractions. It eases integration by
reducing it to the sum of integrals, each of which will most likely give an answer of the form ln(x+a).
And Electrical Engineers doing control problems will use this technique when dealing with transfer
functions.
Example 1)
x 2  3x  1

x 3  6 x 2  11x  6

dx  ? :
x 2  3x  1
x 3  6 x 2  11x  6

x 3  6 x 2  11x  6  x  1x  2 x  3
A
B
C


x 1 x  2 x  3
The first step is to factor the denominator into linear polynomials, if possible. We then use the factors
to set up the second line (A, B, and C are real numbers).
x2  3x  1
A
B
C


x  6 x  11x  6 x  1 x  2 x  3
Ax  2 x  3
B x  1x  3
C x  1x  2 



x  1x  2x  3 x  1x  2x  3 x  1x  2x  3
Ax  2 x  3  B x  1x  3  C x  1x  2 

x3  6 x 2  11x  6
3
2

The RHS of the equation must be manipulated into the form on the third line above. We use the
same technique that we use when adding fractions with unlike denominators: convert each term to its
equivalent with the product of the denominators as the denominator.
x 2  3x  1
x  6 x  11x  6
3
2

Ax  2 x  3  Bx  1x  3  C x  1x  2
x 3  6 x 2  11x  6
 x 2  3x  1  Ax  2 x  3  Bx  1x  3  C x  1x  2 
x  1   12  3 1  1  A 1  2  1  3  1  2 A  A 
1
2
x  2   22  3 2  1  B 2  1 2  3  1   B  B  1
x  3   32  3 3  1  C  3  1 3  2   1  2C  C 
1
2
The next step is to compare the numerators of the original question with the calculated one and solve
for A, B, and C. Method #1 (difficult) is to simplify the RHS of the equation and compare coefficients
thereby setting up three equations in three unknowns (e.g. all coefficients of x 2 terms will sum up to 1,
etc.). Method #2 (easier) is to choose any three values for x and substitute them into both sides of
the equation (e.g. if x=0 then 1=6A+3B+2C) and again obtaining three equations in three unknowns.
Both of these are valid methods but they are time consuming and there is an easier method. Method
#3, which is illustrated, improves on method #2 by selecting specific values of x to eliminate two of
the variables, resulting in three equations with one unknown. Choose the values of x that will equate
the denominator to 0 (i.e. the roots of the polynomial). If x=-1 then the term containing B and C
equates to 0 because they contain (x+1) as a factor. Similarly, x=-2 and x=-3 produce similar results.
1
1
x  3x  1
1
 3
 2 
 2
2
x  6 x  11x  6 x  1 x  2 x  3
2

x 2  3x  1
1 1
1
1
1
dx  
dx  
dx

2 x 1
x2
2 x3
x  6 x  11x  6
1
1

ln x  1  ln x  2   ln x  3  C
2
2
3
2

Note: The numerator polynomial must have a lower degree than the denominator (i.e. a poly of
degree n must have n as the largest exponent of x). To rectify cases that have the degrees equal or
numerator's degree is greater than the denominator's, we must divide using polynomial long division.
The result is a real number plus another fraction that meets the criteria for partial fraction
decomposition.
Another case has to do with a denominator with repeated roots.
Example 2)
x5

dx  ? : x 2  4 x  4  x  22
x  4x  4
x5
A
B


2
x  4 x  4 x  2 x  22
2
The difference here is that we set up two fractions, one with a denominator of x-2 and the other with
(x-2)2.
x5
x  4x  4
2
A
B

x  2 x  22

x  5  Ax  2   B
x2 B 7
x  0  2 A  B  5  2 A  5  7  A  1
In our first example, we had three roots and three variables. In this case, we have one repeated root
and two variables. So we use the root as one value of x and then arbitrarily choose another value of
x. Choose any value you want, but remember that you have to do the calculations and hence the
reasoning for x=0. If x=0 then both A and B will be in the equation. Since we know B, it is a simple
substitution to solve for A.
x5
x  4x  4
2


1
7

x  2 x  2 2
x5
x  4x  4
2
 ln x  2 
dx  
1
1
dx  7 
dx
x2
x  22
7
C
x2
If you cannot factor the denominator into factors of the form x-a where a is a real number, then you
can still use a modified version of this technique. Please see PFDExamples2 for an illustration.