5.2. PARTIAL FRACTIONS 163
Problem 5.15 Perform the following integrals.
1.
Z
q
sin
2
.x/ C 1 cos.x/ dx
2.
Z
e
3x
p
e
2x
C 4 dx
3.
Z
.x C1/
p
x
2
C 2x C2 dx
4.
Z
p
1 e
2x
dx
5.
Z
p
1 tan
2
.x/
tan.x/
sec
2
.x/ dx
6.
Z
x
p
1 x
2
p
1 Cx
2
dx
Problem 5.16 Compute
Z
x
P
sin
1
.x/ dx
5.2 PARTIAL FRACTIONS
is section introduces a novel algebraic technique for setting up integrals. e first thing we
need is a useful fact about polynomials over the real numbers.
Knowledge Box 5.1
Factorization of polynomials
e only polynomials that do not factor over the real numbers are
quadratics with no roots. Any polynomial of degree 3 or more factors
into linear polynomials and quadratics that do not factor.
What can we do with this fact? Let’s look at an example to set the stage.
Example 5.17 Compute
Z
dx
x
2
3x C2
Solution:
Notice that
1
x 2
1
x 1
D
.x 1/ .x 2/
.x 1/.x 2/
D
1
x
2
3x C2
164 5. METHODS OF INTEGRATION II
is means that:
Z
dx
x
2
3x C2
D
Z
dx
x 2
Z
dx
x 1
D ln jx 2j ln jx 1j C C
e problem with this solution is that “notice” is not a recognized algebra technique. In order to
gain the ability to “notice” in this fashion we need one additional fact.
Equal polynomials have
equal coefficients.
Suppose that
A
x 1
C
B
x 2
D
1
.x 1/.x 2/
en, if we cross-multiply on the left side of the equation, we get
A.x 2/ C B.x 1/
.x 1/.x 2/
D
1
.x 1/.x 2/
which simplifies to
.A CB/x 2A B D 1 D 0x C 1
Since these polynomials are equal we get the system of equations:
A CB D 0
2A B D 1
A D 1 Second equation plus first.
B D 1 Substitute AD 1 into the first equation.
So A D 1 and B D 1 tell us the thing we noticed:
1
x 2
1
x 1
D
1
x
2
3x C2
˙
is is the basis of integration by partial fractions. e fact that a polynomial factors into
linear and unfactorable quadratic terms means that we can always break a rational function into
(i) polynomial terms and (ii) rational functions of the form
A
x u
or
Bx C C
x
2
C sx Ct
Integration by partial fractions is used to integrate rational functions.
5.2. PARTIAL FRACTIONS 165
Knowledge Box 5.2
Steps of integration by partial fractions
1. If the numerator of the rational function is not lower degree than the denom-
inator, divide to transform the problem into a polynomial plus a rational
function with a higher degree denominator.
2. Factor the denominator.
3. Set the rational function equal to the sum of partial fraction terms.
4. Clear the denominator by cross multiplication.
5. Solve for the coefficients of the partial fraction terms based on the equal poly-
nomials resulting from cross multiplication.
6. Perform the resulting integrals.
If a factor of the denominator is repeated, it gets one partial fraction term for each power of the
denominator. Let’s work some examples.
Example 5.18 Find
Z
dx
x
2
4x C 3
Solution:
e numerator is already lower degree. Factor the denominator. x
2
4x C 3 D .x 3/.x 1/
which gives us
1
x
2
4x C 3
D
A
x 1
C
B
x 3
meaning that
1
x
2
4x C 3
D
A.x 3/ C B.x 1/
x
2
4x C 3
yielding the polynomial equality 1 D .A C B/x .3A C B/. We get two equations – one for
the coefficient of x and one for the constant term:
166 5. METHODS OF INTEGRATION II
3A CB D 1 for constant term
A CB D 0 for coefficient of x
A D B
2B D 1
B D 1=2
A D 1=2
So
1
x
2
4x C 3
D
1
2
1
x 3
1
x 1
Setting up the integral:
Z
dx
x
2
4x C 3
D
1
2
Z
1
x 3
1
x 1
dx
D
1
2
.
ln jx 3jln jx 1j
/
C C
˙
Example 5.19 Compute
Z
2x C5
x
2
x
Solution:
e numerator is already lower degree. Factor the denominator and set up the partial
fractions.
2x C5
x.x 1/
D
A
x
C
B
x 1
Cross multiply and extract the polynomials equations.
2x C5 D A.x 1/ CBx
which gives us the linear system
A CB D 2
5 D A
A D 5
B D 7
5.2. PARTIAL FRACTIONS 167
So
Z
2x C5
x
2
x
D
Z
7
x 1
dx
Z
5
x
dx
D 7 ln jx 1j 5 ln jxj C C
˙
e next example will be our first with a repeated root, letting us showcase the fact that partial
fractions have one part for each point of multiplicity of a root.
Example 5.20 Compute
Z
dx
x
3
x
2
Solution:
We begin, as always, by factoring: x
3
x
2
D x
2
.x 1/. e root x D 0 has multiplicity
two. is means that both x and x
2
are factors. So, the partial fraction decomposition is:
1
x
3
x
2
D
A
x
C
B
x
2
C
C
x 1
So, once we clear the denominator (D x
2
.x 1/) we get
1 D Ax.x 1/ C B.x 1/ C Cx
2
or
.A CC /x
2
C .B A/x B D 0x
2
C 0x C1
e resulting system of equations is:
A CC D 0
B A D 0
B D 1
So B D 1, A D 1, and C D 1. Now we can integrate:
Z
dx
x
3
x
2
D
Z
1
x
1
x
2
C
1
x 1
dx
D ln jxj C
1
x
C ln jx 1jC C
˙
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