218
4.8
4.8.1
CHAPTER 4. INTEGRALS
Partial Fraction Decomposition
Need to Know
The following material is assumed to be known for this section. If this is not
the case, you will need to review it.
1. When are two polynomials equal?
2. Relationship between the roots (zeros) of a polynomial and its factors.
3. How to factor polynomials.
4. Long division of polynomials.
4.8.2
Introduction
This handout describes a method to rewrite a fraction we do not know how
to integrate into simpler (partial) fractions we know how to integrate. This
method works for rational functions, that is functions which can be written as
the quotient of two polynomial functions.
For the remaining part of this document, we will assume that we have a
p (x)
in which degree of p (x) < degree of q (x). If this is not
rational function
q (x)
the case, we can always perform long division. For example, if we were given
x4
. We would perform long division to obtain
the fraction 2
x
1
x4 = x2
x4
x2
1
1
= x2 + 1 +
x2 + 1 + 1
1
x2
1
1
.
x2 1
In this class, we will use partial fraction decomposition as an integration
technique. The ultimate goal is to decompose a fraction so we can integrate it.
In this document, we will focus on the decomposition. Keep in mind why we
are doing this decomposition.
How we perform partial fraction decomposition depends of the denominator
of the fraction. We consider several cases.
We then would apply partial fraction decomposition to
4.8.3
Case 1: q(x) is a product of distinct linear factors.
Let us assume that q (x) is a product of n distinct linear factors that is
q (x) = (a1 x + b1 ) (a2 x + b2 ) ::: (an x + bn )
Then,
p (x)
A1
A2
An
=
+
+ ::: +
q (x)
a1 x + b1
a2 x + b2
an x + bn
4.8. PARTIAL FRACTION DECOMPOSITION
219
Finding the decomposition amounts to …nding the coe¢ cients A1 , ..., An . Hence,
we have n coe¢ cients to …nd. It is a known fact in mathematics that to …nd
n unknowns, we need n equations (as many equations as unknowns). Thus, we
need to generate n equations involving the coe¢ cients A1 , ..., An . Solving these
equations will give us the coe¢ cients we want. This can be done two di¤erent
ways. We illustrate this in the examples below.
Remark 319 p (x) does not play a role in the way the decomposition is written.
x2 + 1
?
(x 1) (x 3) (x + 1)
When the question is asked this way, you only have to write the decomposition
but not to …nd the coe¢ cients. From what is explained above, we see that
Example 320 Write the decomposition of
(x
x2 + 1
A
B
C
=
+
+
1) (x 3) (x + 1)
x 1 x 3 x+1
x
x2 + 2x
We begin by factoring the denominator. We obtain
Example 321 Find the decomposition for
3
x
(x + 3) (x
1)
. According
to the decomposition we wrote above, we have:
x
(x + 3) (x
=
1)
A
B
+
x+3 x 1
We need to …nd A and B. Multiplying each side by the denominator of the
fraction on the left and simplifying, we obtain:
x = A (x
= Ax
1) + B (x + 3)
A + Bx + 3B
= (A + B) x
A + 3B
Two polynomials are equal if their corresponding coe¢ cients are equal. This
gives us the following system:
A+B =1
A + 3B = 0
8
>
< A= 3
4
The solution of this system is:
1
>
: B=
4
x2
x
+ 2x
3
=
1
4
Thus, we have
3
1
+
x+3 x 1
There is an easier way which works in this case. We show it on the next example.
220
CHAPTER 4. INTEGRALS
x2 x
(x 1) (x + 5) (x
The denominator is already factored. The decomposition is:
Example 322 Find the decomposition for
(x
x2 x
1) (x + 5) (x
3)
=
A
x
1
+
3)
B
C
+
x+5 x 3
We need to …nd A, B, and C . We begin the same way, we multiply each side
by the denominator of the fraction on the left, and simplify. We obtain:
x2
x = A (x + 5) (x
3) + B (x
1) (x
3) + C (x
1) (x + 5)
Then we notice that since the above equality is true for every x, it will be true
for speci…c values of x. We select values for x which will make all but one of
the coe¢ cients go away. We will then be able to solve for that coe¢ cient. More
precisely,
When x = 1, we obtain:
0 = A (6) ( 2)
A=0
When x =
5, we obtain
30 = B ( 6) ( 8)
30
B=
48
5
B=
8
When x = 3, we obtain
6 = C (2) (8)
3
C=
8
Therefore, the decomposition is:
(x
4.8.4
x2 x
1) (x + 5) (x
3)
=
1
8
5
3
+
x+5 x 3
Case 2: q(x) is a product of linear factors, some
being repeated
The factors which are not repeated will be decomposed as above. Suppose
m
that q (x) also contains (ax + b) that is ax + b is repeated m times. The
decomposition for this factor will be
A2
A1
Am
+
+ ::: +
m
ax + b (ax + b)2
(ax + b)
4.8. PARTIAL FRACTION DECOMPOSITION
Example 323 Find the decomposition for
221
x2
(x
2
2) (x + 1)
3
The decomposition is:
x2
(x
2
2) (x + 1)
3
=
A
x
2
+
D
B
C
+
+
3
x + 1 (x + 1)2
(x + 1)
Then, we proceed as before. We multiply each side by the denominator of the
fraction on the left and simplify. We obtain:
x2
3
2 = A (x + 1) + B (x
2
2) (x + 1) + C (x
2) (x + 1) + D (x
2)
We need to …nd A, B, C, and D. We can use either of the methods described
in the …rst case.
If x = 2, we get
2 = 27A
2
A=
27
If x =
1, we get
1=
3D
1
D=
3
We still have to …nd B and C. For this, we choose two more values for x
and write the corresponding system. We now know A and D, so we can
use the value we found for them.
If x = 1, we get
1 = 8A
16
=
27
16
4B + 2C =
27
34
4B + 2C =
27
4B
2C
4B
2C
D
1
3
9
27
+
27 27
If x = 0, we get
2 = A 2B 2C 2D
2
2
=
2B 2C
27
3
2
18 54
2B + 2C =
+
27 27 27
38
2B + 2C =
27
222
CHAPTER 4. INTEGRALS
Thus, we need to solve
The solution is :
x2
(x
4.8.5
B=
2
2) (x + 1)
3
=
8
>
< 4B + 2C = 34
27
38
>
: 2B + 2C =
27
2
27 ; C
=
2
27 (x 2)
7
9
. Putting all this together, we get
1
2
7
+
+
2
3
27 (x + 1) 9 (x + 1)
3 (x + 1)
Case 3: q(x) is a product of distinct irreducible
quadratic factors
Recall that a term is called irreducible if it cannot be factored any further. Thus
2
x2 + x + 1 is irreducible, so is x2 + 1. Be careful, (x + 1) is not considered a
quadratic term. You must think of it as a linear term appearing twice. We
can generalize what we did in the previous two cases as follows. Instead of
thinking of linear factors, think that when we write the decomposition, the
degree of the term in the numerator is 1 less than the degree of the term in the
denominator. When we had linear factors in the denominator, it meant that we
had to have terms of degree 0 in the numerator, that is we had constant terms.
If the denominator contains irreducible quadratic factors, then the numerator
will contain linear terms. We look at an example to see how the decomposition
is written.
x
+ 1) (x2 + 2)
If we apply a method similar to that of case 1, the decomposition will contain
two fractions, one for each irreducible factor. However, since the degree of the
denominator is now 2, the degree of the numerator will be 1, that is we will have
linear terms. Recall that a linear term is of the form ax + b. Thus,
Example 324 Find the decomposition for
(x2
Ax + B
Cx + D
x
= 2
+ 2
(x2 + 1) (x2 + 2)
x +1
x +2
To solve, we proceed as above. First, we multiply each side by the denominator
of the fraction on the left, and simplify. We obtain:
x = (Ax + B) x2 + 2 + (Cx + D) x2 + 1
We then pick 4 di¤ erent values for x to get a system of 4 equations, which we
solve. The answer is:
(x2
x
x
= 2
2
+ 1) (x + 2)
x +1
x2
x
+2
4.8. PARTIAL FRACTION DECOMPOSITION
4.8.6
223
Case 4: q(x) is a product of irreducible quadratic
factors, some being repeated
This is similar to case 2, with linear terms in the numerator and quadratic terms
in the denominator.
Example 325 Write the decomposition for
2x
1
(x2 + x + 1)
3
=
2x
1
(x2 + x + 1)
3
Ax + B
Ex + F
Cx + D
+
+
3
x2 + x + 1 (x2 + x + 1)2
(x2 + x + 1)
We would …nd the coe¢ cients as above.
4.8.7
General Case: q(x) is a mixture of the above
Example 326 Find the decomposition for
2x
(x
1)
2
(x2
1
2x
(x
2
1)
(x2
1
+ x + 1)
3
A
B
+
2
x
1
+ x + 1)
(x 1)
Cx + D
Ex + F
Gx + H
+ 2
+
+
3
2
x + x + 1 (x2 + x + 1)2
(x + x + 1)
3
=
The answer is
2x
(x
4.8.8
1)
2
(x2
1
+ x + 1)
3
=
1
2
27 (x 1)
x+1
1
+
2
27 x + x + 1
1
27 (x 1)
1
9 (x2
+ x + 1)
2
1
x+3
2
3 (x + x + 1)3
Application
The idea behind this decomposition is that once the fraction is decomposed, we
can integrate it.
Z
x
Example 327 Find
dx
2
x + 2x 3
x
1
3
1
=
+
. Therefore,
We found earlier that 2
x + 2x 3
4 x+3 x 1
Z
Z
Z
x
1
dx
dx
dx
=
3
+
2
x + 2x 3
4
x+3
x 1
3
1
= ln jx + 3j + ln jx 1j
4
4
Using substitution
224
CHAPTER 4. INTEGRALS
Z
x
dx
+ 1) (x2 + 2)
x
We saw earlier as an example that the partial fraction decomposition of 2
(x + 1) (x2 + 2)
was
x
x
x
= 2
2
2
2
(x + 1) (x + 2)
x +1 x +2
Example 328 Find
(x2
Therefore
Z
x
dx =
(x2 + 1) (x2 + 2)
Z
Z
x
dx
x2 + 1
x
dx
x2 + 2
We can do the …rst integral by substitution. If we let u = x2 +1, then du = 2xdx
and therefore
Z
Z
1 du
x
dx
=
x2 + 1
2
u
1
= ln juj
2
1
= ln x2 + 1
2
The second integral is done in a similar way. We obtain
Z
x
1
dx = ln x2 + 2
x2 + 2
2
It follows that
Z
x
1
1
dx = ln x2 + 1
ln x2 + 2
(x2 + 1) (x2 + 2)
2
2
1
x2 + 1
= ln
2
x2 + 2
s
x2 + 1
= ln
+C
x2 + 2
Example 329 Find
R
x3
dx
x2 1
The function we are integrating is a rational function. However, the degree of
the numerator is greater than or equal to the degree of the denominator. So, the
…rst step is to perform long division. We obtain
x3
x2
It follows that
Z
x3
x2
1
1
dx =
=x+
Z
x
x2
xdx +
1
Z
x
x2
1
dx
(4.11)
4.8. PARTIAL FRACTION DECOMPOSITION
225
We can do the …rst integral. The second, is the integral of a rational function.
x
To be able to evaluate it, we …rst decompose 2
into partial fractions.
x
1
x
x2
1
x
(x 1) (x + 1)
B
A
+
=
x 1 x+1
=
We need to …nd A and B. We do it using the techniques described above. First,
we multiply each side by the denominator of the fraction on the left to obtain
x = A (x + 1) + B (x
1)
1
When x = 1, we get 1 = 2A or A = . When x =
2
1
B = . Therefore,
2
1
1
1
x
=
+
x2 1
2 x 1 x+1
1, we get
1=
2B or
If we replace what we just found in equation 4.11, we obtain
Z
Z
Z
Z
x3
1
dx
1
dx
dx
=
xdx
+
+
2
x
1
2
x 1 2
x+1
These are integrals we can handle.
Z
x3
x2
1
1
dx
=
+ ln jx 1j + ln jx + 1j
2
x
1
2
2
2
1
x2
+ ln x2 1
=
2
2
4.8.9
Problems
Do the problems below:
1. Write the partial fraction decomposition of
ing the coe¢ cients.
x2
2. Write the partial fraction decomposition of
the coe¢ cients.
R
x 9
3. Find
(x + 5) (x
4. Find
5. Find
R3
2
R
1
x2
(x
1
2)
dx
dx
10
dx
1) (x2 + 9)
2
+ 3x
4
without evaluat-
x 1
without evaluating
x3 + x2
226
CHAPTER 4. INTEGRALS
R
6. Find
6
dx
R x3 + 4
dx
x2 + 4
Z 2
4x 1
dx
(x
1) (x + 2)
4
Z
x2
dx
x+1
Z 2
x +1
dx
x2 x
Z
dx
where a 6= 0
x2 a2
Z 1
2x + 3
2 dx
0 (x + 1)
Z
dx
7. Find
8. Find
9. Find
10. Find
11. Find
12. Find
13. Find
4.8.10
x
x
2
(x
1) (x + 4)
Answers
1.
2
x2 + 3x
2.
x 1
A
B
C
= + 2+
x3 + x2
x
x
x+1
3.
4.
5.
6.
R
x 9
(x + 5) (x
R3
2
R
R
4
1
x2
(x
1
2)
dx =
1
2
dx = 2 ln jx + 5j
ln 3
1
2
6
dx = x + 6 ln jx
ln jx
2j + C
ln 2
1
10
dx = tan
1) (x2 + 9)
3
x
x
A
B
+
x+4 x 1
=
1
x
3
1
ln x2 + 9 + ln jx
2
1j + C
6j + C
R x3 + 4
1
x
1
dx = tan 1
2 ln x2 + 4 + x2 + C
x2 + 4
2
2
2
Z 2
4x 1
2
8.
dx = 3 ln 4 ln 3 3 ln 6 = 3 ln 2 4 ln 3 = ln 3+3 ln
(x
1)
(x
+
2)
3
4
4x 1
1
3
The partial fraction decomposition of
is
+
.
(x 1) (x + 2)
x 1 x+2
7.
4.8. PARTIAL FRACTION DECOMPOSITION
9.
Z
x2
dx = 12 x2
x+1
x + ln jx + 1j + C
The partial fraction decomposition of
10.
Z
x2 + 1
dx = x
x2 x
227
ln x + 2 ln jx
x2
is x
x+1
1+
1
.
x+1
1j + C
1
2
x2 + 1
is 1
+
because
x2 x
x
x 1
2
x +1
x+1
x+1
2
1
= 2
+ 1 and the decomposition of 2
is
.
x2 x
x
x
x
x
x 1 x
Z
dx
11.
= where a 6= 0
2
x
a2
R
dx
1
1
=
ln jx aj
ln ja + xj + C
2
2
x
a
2a
2a
1
1
1
is
.
The partial fraction decomposition of 2
x
a2
2a (x a) 2a (a + x)
Z 1
2x + 3
1
12.
2 dx = 2 + 2 ln 2
0 (x + 1)
1
2
2x + 3
The partial fraction decomposition of
2 is
2 + x + 1.
(x + 1)
(x + 1)
Z
1
1
1
dx
ln jx 1j +
ln jx + 4j + C
13.
=
2
5
(x
1)
25
25
(x 1) (x + 4)
1
1
The partial fraction decomposition of
is
2
2
(x 1) (x + 4)
5 (x 1)
1
1
+
.
25 (x 1) 25 (x + 4)
The partial fraction decomposition of