Roberto’s Notes on Integral Calculus
Chapter 2: Integration methods
Section 15
Integration by partial fractions
with non-repeated quadratic factors
What you need to know already:
 How to use the integration method of partial
fractions when the denominator is a product of
non-repeated linear factors.
I hope you will be comforted by the fact that applying the method
of partial fractions in the case when the denominator is a product of
non-repeated, irreducible quadratic factors is only a small modification
of what you saw happening with non-repeated linear factors.
Strategy for splitting a proper
fraction whose denominator consists
of non-repeated irreducible
quadratic factors
f  x 
Integral Calculus
a p x  a p 1 x
x
2
 b1 x  c1
p 1
 a1 x  a0
 x
2
 br x  cr
 How to use the integration method of partial
fractions when the denominator is a product of
non-repeated (irreducible) quadratic factors.
where all quadratic factors are different from each
other, then it can be split into a sum of the form:
f  x 
h1 x  k1

x  b1x  c1
2

hr x  kr
x  br x  cr
2
for appropriate values of the coefficients hi and ki .
In other words, we can split it as in the linear case,
but using linear numerators, instead of constants.
And how do we get these coefficients? Do we still have smart values available?
When a rational function in proper form can be
written as:
p
What you can learn here:
Unfortunately not, since these factors are irreducible and therefore have no
roots. Instead we do the following.

Chapter 2: Integration methods
Section 15 Partial fractions with non-repeated quadratic factors
Page 1
Strategy for finding the partial
fractions coefficients in the presence
of irreducible quadratic factors
To get the needed values of the coefficients we
follow the same procedure as with linear factors, but
since there are no smart values to use, we use easy
values, that is, small and simple values that make
computations easy, such as 0, 1, or other values
suggested by the particular case.
x  1  2  2  a  b   10  c  d 
 1  a  b  5c  5d
x  1  0  2  a  b   10  c  d 
 0  a  b  5c  5d
x  2  1  5  2a  b   13 2c  d 
 1  10a  5b  26c  13d
These four equations together form a linear system that can be solved by
elimination and back substitution (long way) or by using linear algebra
methods. With the latter, and the help of a calculator, we find that:
ab
An example is badly needed, so here it is:
Therefore:
x 1
Example: f  x   2
x  9 x2  1


1
1
; cd 
8
8


Notice that the denominator is already factored, something that will occur
fairly frequently, unless the factoring is extremely simple, so we apply the
procedure:
x 1
1
1
1 





2
2
x2  9 x2  1 8  x  9 x  1 


Now what? How do we integrate the partial fractions?
That is another step where things can get complicated.
We equate the given fraction to a sum of fractions with a single factor in the
denominator and a linear numerator. We can also use an easier notation for
the constants:

x 1
ax  b cx  d
 2
 2
2
x  9 x 1 x  9 x 1
2


Next we combine the fractions on the right side and equate the resulting
numerators:



x  1   ax  b  x 2  1   cx  d  x 2  9

Since we have no smart values, we use some simple, easy values.
x  0  1  b  9d
Integral Calculus
Chapter 2: Integration methods
Section 15 Partial fractions with non-repeated quadratic factors
Page 2
Strategy for integrating a single
proper fraction with an irreducible
quadratic in the denominator
If the quadratic denominator includes a first degree
term, we must complete the square before setting up
the substitution.
If the numerator of the fraction consists of the
constant only, an arctangent will be obtained as
integral, possibly after a simple substitution.
Example:

In order to integrate this one, we first need to place the denominator in
completed square form:

2x  3
dx 
x  4x  5
2

2x  3
dx 
x  4x  4  4  5
2

We now notice that the derivative of the denominator is
2x  3
x  2  1
2
dx
2  x  2  2 x  4 ,
so we adjust the numerator to show this expression and split the resulting
fraction:
If the numerator of the fraction consists of the linear
term only, a substitution will allow us to use the
reciprocal rule, thus obtaining a logarithm.
If the numerator of the fraction includes both terms
of the linear function, the fraction may have to be
split further, thus involving both a logarithm and an
arctangent.
2x  3
dx
x2  4 x  5


2x  3
x  2  1

2
dx 
2x  4
x  2  1
2


2x  4  7
x  2  1
dx  7
2
dx
1
x  2  1
2
dx
The first integral can be computed with the substitution u   x  2   1 ,
2
while the second can be done with the substitution v  x  2 . In this way
we get two basic integrals leading to (check the details!):

Example:
Since

x 1

x 1
dx
2
x  9 x2  1


1
1
1 
  2
 2  , we can write:
x  9 x 1 8  x  9 x 1
2

2


1

dx

8 
x2  9 x2  1
x 1



dx

2
x 9

dx 

x 2  1 
2x  3
x  2  1
2


dx  ln  x  2   1  7 arctan  x  2   c
2
As you can see, there can be quite a bit of work involved, but the
procedure works. Need I say that practise is needed to become
proficient in the method? Well, I just said it, didn’t I! And if you
don’t see enough questions here, look at the next section on the
general case, as well as anything you can find in your resources.
By using the substitution 3u  x in the first integral, we obtain:
1 1
x

   tan 1  tan 1 x   c
8 3
3

Integral Calculus
Chapter 2: Integration methods
Section 15 Partial fractions with non-repeated quadratic factors
Page 3
Summary
 To split a rational function with non-repeated irreducible quadratic factors in the denominator, we use the same method as for linear factors, but we need to allow for lnear
numerators.
 To find the coefficients of the numerators, we must use easy values, sine no smart values are available.
Common errors to avoid
 Watch the algebra, as the process can be quite long, and practice lots to gain familiarity.
Learning questions for Section I 2-15
Review questions:
1. Explain, in your own words, how to integrate a proper rational function whose denominator is a product of non-repeated quadratic factors.
Memory questions:
1. In the method of partial fractions, what is needed in the numerators of the smaller fractions when the denominator is irreducible quadratic?
Computation questions:
Compute the integrals proposed in questions 1-4.
1.

7  6x
dx
x  2x  3
2.
2
Integral Calculus
Chapter 2: Integration methods

x2
dx
x  2x  3
2
Section 15 Partial fractions with non-repeated quadratic factors
Page 4
3.

2x  5
dx
x x4
4.
2

2x  5
dx
x  x  4 x2  2
2


Theory questions:
1. When do we need to complete the square within the method of partial fractions?
2. Which two non-rational functions may occur in the integral of a rational function?
1. Construct a simple proper rational function whose denominator consists of one irreducible quadratic factor or two such factors multiplied together. Then integrate it.
What questions do you have for your instructor?
Integral Calculus
Chapter 2: Integration methods
Section 15 Partial fractions with non-repeated quadratic factors
Page 5
Integral Calculus
Chapter 2: Integration methods
Section 15 Partial fractions with non-repeated quadratic factors
Page 6