168 5. METHODS OF INTEGRATION II
Next we look at a shortcut that permits us to, in some cases, avoid having to solve the system of
equations for the coefficients of the partial fraction decomposition. Instead, we plug the roots
into the equations. May terms zero out and, often, we get the values of the partial fraction
coefficients directly.
Example 5.21 Compute
Z
dx
x
3
6x C11x 6
Solution:
Start by factoring: x
3
6x C11x 6 D .x 1/.x 2/.x 3/ which gives us a partial
fractions form:
1
x
3
6x C11x 6
D
A
x 1
C
B
x 2
C
C
x 3
Clearing the denominator we get
1 D A.x 2/.x 3/ CB.x 1/.x 3/ C C.x 1/.x 2/
which sets us up for the shortcut:
1 D A.1/.2/ D 2A When x D 1
1 D B.1/.1/ D B When x D 2
1 D C.2/.1/ D 2C When x D 3
So A D 1=2, B D 1, and C D 1= 2. Integrate:
Z
dx
x
3
6x C11x 6
D
Z
1=2
x 1
1
x 2
C
1=2
x 3
dx
D
1
2
ln jx 1jln jx 2j C
1
2
ln jx 3jCC
˙
What happened? Let’s summarize our findings in a Knowledge Box.
5.2. PARTIAL FRACTIONS 169
Knowledge Box 5.3
Finding the coefficients of a partial fractions decomposition
• Equal polynomials have equal coefficients for equal power terms.
• Equal polynomials are equal for any specific value of their variable.
e shortcut for finding the partial fraction constants is to plug in values of x that zero out
all but one term. Warning: this isn’t always possible. Terms with multiplicity above one or non-
factorable quadratic terms can mess up the shortcut. e next set of examples involve integrating
reciprocals of quadratics that don’t factor. ese will arise as part of partial fraction decomposi-
tions.
Example 5.22 Compute
Z
dx
x
2
C 4x C 5
Solution:
e denominator doesn’t factor. Let’s try completing the square.
Z
dx
x
2
C 4x C 5
D
Z
dx
x
2
C 4x C 4 C 1
D
Z
dx
.x C2/
2
C 1
Let u D x C 2, so du D dx
D
Z
du
u
2
C 1
D tan
1
.u/ CC
D tan
1
.x C2/ C C
˙
170 5. METHODS OF INTEGRATION II
Knowledge Box 5.4
Integrating the reciprocal of unfactorable quadratics
1. Complete the square to place the quadratic in the form a.x b/
2
C c
2.
1
a.x b/
2
C c
D
1
c
a
c
.x b/
2
C 1
D
1
c
r
a
c
.x b/
2
C 1
!
3. Let u D
r
a
c
.x b/ so that du D
r
a
c
dx and dx D
r
c
a
du
4.
Z
1
a.x b/
2
C c
dx D
1
c
r
c
a
tan
1
.u/ C C
D
1
p
ac
tan
1
r
a
c
.x b/
C C
With this particular type of u-substitution in place we can do the next batch of integrations by
partial fractions.
Example 5.23 Compute
Z
dx
x
3
C x
Solution:
Factor and we see that x
3
C x D x.x
2
C 1/: So the partial fraction form is:
1
x
3
C x
D
A
x
C
Bx C C
x
2
C 1
We need Bx C C in the numerator of the second part instead of just B because x
2
C 1 is
quadratic. Clearing the denominator we get the polynomial equation:
1 D A.x
2
C 1/ C x.Bx C C /
Plug in x D 0 and we see that 1 D A. is tells us that
1 D .x
2
C 1/ C Bx
2
C Cx
or
x
2
C 0x D Bx
2
C Cx
5.2. PARTIAL FRACTIONS 171
so B D 1 and and C D 0. We are now prepared to integrate
Z
dx
x
3
C x
D
Z
1
x
x
x
2
C 1
dx
D ln jxj
1
2
ln jx
2
C 1jC C
e second part of the integral is a u-substitution with u D x
2
C 1; we have done it before.
˙
Let’s try a less neat integral of this sort.
Example 5.24 Compute
Z
3x dx
x
3
9x
2
C 25x 25
Solution:
Factor, and we get x
3
9x
2
C 25x 25 D .x 5/.x
2
4x C 5/ D .x 5/..x 2/
2
C 1/.
at makes the partial fraction form
3x dx
x
3
9x
2
C 25x 25
D
A
x 5
C
Bx C C
x
2
4x C 5
Clear the denominator and we get
0x
2
C 3x C0 D A.x
2
4x C 5/ C .Bx C C /.x 5/ D Ax
2
4Ax C5A C Bx
2
5Bx CC x 5C
yielding the simultaneous equations
A CB D 0
4A 5B C C D 3
5A 5C D 0
So, A D B, A D C , and 4A C 5A CA D 2A D 3. So, A D C D 3=2 and B D 3=2.
172 5. METHODS OF INTEGRATION II
is permits us to integrate:
Z
3x dx
x
3
9x
2
C 25x 25
D
3
2
Z
1
x 5
x 1
.x 2/
2
C 1
dx
D
3
2
Z
1
x 5
x 2
.x 2/
2
C 1
C
1
.x 2/
2
C 1
dx
Let u D x 2 with du D dx for the second and third terms
D
3
2
ln jx 5j
1
2
ln.u
2
C 1/ C tan
1
.u/
C C
D
3
2
ln jx 5j
1
2
ln.x
2
4x C 5/ C tan
1
.x 2/
C C
˙
Example 5.25 Compute
Z
dx
x
3
1
Solution:
Factor: x
3
1 D .x 1/.x
2
C x C1/. en
1
x
3
1
D
A
x 1
C
Bx C C
x
2
C x C1
Clear the denominator and we get
1 D A.x
2
C x C1/ C .Bx C C /.x 1/
yielding:
A CB D 0
A B C C D 0
A C D 1
Solving the linear system we get A D 1=3, B D 1=3, C D 2=3.
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