856
CHAPTER 7
SOLUTION
TECHNIQUES OF INTEGRATION
Let t.y/ D sinh!1 .tan y/. Then
t 0 .y/ D
For 0 " y <
!
2,
cos2 y
p
1
tan2 y
C1
D
cos2 y
1
q
1
cos2 y
D
1
cos2 y !
1
j cos yj
D
1
D j sec yj:
j cos yj
sec y > 0; therefore t 0 .y/ D sec y. Integrating this last relation yields
Z y
1
t.y/ # t.a/ D
dt:
cos
t
a
For this to be of the desired form, we must have t.a/ D sinh!1 .tan a/ D 0. The only value for a that satisfies this equation is
a D 0.
49. The relations cosh.it/ D cos t and sinh.it/ D i sin t were discussed in the Excursion. Use these relations to show that the
identity cos2 t C sin2 t D 1 results from setting x D it in the identity cosh2 x # sinh2 x D 1.
SOLUTION
Let x D it. Then
cosh2 x D .cosh.it//2 D cos2 t
and
sinh2 x D .sinh.it//2 D i 2 sin2 t D # sin2 t:
Thus,
1 D cosh2 .it/ # sinh2 .it/ D cos2 t # .# sin2 t/ D cos2 t C sin2 t;
as desired.
7.5 The Method of Partial Fractions
PreliminaryR Questions
1. Suppose that
f .x/ dx D ln x C
p
x C 1 C C . Can f .x/ be a rational function? Explain.
SOLUTION No, f .x/ cannot be a rational function because the integral of a rational function cannot contain a term with a nonp
integer exponent such as x C 1.
2. Which of the following are proper rational functions?
x
(a)
x#3
x 2 C 12
(c)
.x C 2/.x C 1/.x # 3/
(b)
(d)
4
9#x
4x 3 # 7x
.x # 3/.2x C 5/.9 # x/
SOLUTION
(a)
(b)
(c)
(d)
No, this is not a proper rational function because the degree of the numerator is not less than the degree of the denominator.
Yes, this is a proper rational function.
Yes, this is a proper rational function.
No, this is not a proper rational function because the degree of the numerator is not less than the degree of the denominator.
3. Which of the following quadratic polynomials are irreducible? To check, complete the square if necessary.
(a) x 2 C 5
(b) x 2 # 5
(c) x 2 C 4x C 6
(d) x 2 C 4x C 2
SOLUTION
(a)
(b)
(c)
(d)
Square is already completed; irreducible.
p
p
Square is already completed; factors as .x # 5/.x C 5/.
x 2 C 4x C 6 D .x C 2/2 C 2; irreducible.
p
p
x 2 C 4x C 2 D .x C 2/2 # 2; factors as .x C 2 # 2/.x C 2 C 2/.
4. Let P .x/=Q.x/ be a proper rational function where Q.x/ factors as a product of distinct linear factors .x # ai /. Then
Z
P .x/ dx
Q.x/
(choose the correct answer):
(a) is a sum of logarithmic terms Ai ln.x # ai / for some constants Ai .
(b) may contain a term involving the arctangent.
S E C T I O N 7.5
SOLUTION
The Method of Partial Fractions
857
The correct answer is (a): the integral is a sum of logarithmic terms Ai ln.x # ai / for some constants Ai .
Exercises
1. Match the rational functions (a)–(d) with the corresponding partial fraction decompositions (i)–(iv).
x 2 C 4x C 12
2x 2 C 8x C 24
(a)
(b)
2
.x C 2/.x C 4/
.x C 2/2 .x 2 C 4/
(c)
x 2 # 4x C 8
.x # 1/2 .x # 2/2
(d)
x 4 # 4x C 8
.x C 2/.x 2 C 4/
4
4x # 4
#
x C 2 x2 C 4
#8
4
8
5
(ii)
C
C
C
2
x#2
x #1
.x # 2/
.x # 1/2
1
2
#x C 2
1
4
(iii)
C
C 2
(iv)
C 2
2
xC2
x
C
2
.x C 2/
x C4
x C4
(i) x # 2 C
SOLUTION
x 2 C 4x C 12
1
4
D
C 2
.
2
xC2
.x C 2/.x C 4/
x C4
2x 2 C 8x C 24
1
2
#x C 2
(b)
D
C
C 2
.
xC2
.x C 2/2 .x 2 C 4/
.x C 2/2
x C4
x 2 # 4x C 8
#8
4
8
5
(c)
D
C
C
C
.
x#2
x#1
.x # 1/2 .x # 2/2
.x # 2/2
.x # 1/2
x 4 # 4x C 8
4
4x # 4
(d)
Dx#2C
#
.
x C 2 x2 C 4
.x C 2/.x 2 C 4/
(a)
2. Determine the constants A; B:
2x # 3
A
B
D
C
.x # 3/.x # 4/
x#3
x#4
SOLUTION
Clearing denominators gives
2x # 3 D A.x # 4/ C B.x # 3/:
Setting x D 4 then yields
8 # 3 D A.0/ C B.1/
or
B D 5;
while setting x D 3 yields
or
6 # 3 D A.#1/ C 0
A D #3:
3. Clear denominators in the following partial fraction decomposition and determine the constant B (substitute a value of x or use
the method of undetermined coefficients).
3x 2 C 11x C 12
1
B
3
D
#
#
x C 1 x C 3 .x C 3/2
.x C 1/.x C 3/2
SOLUTION
Clearing denominators gives
3x 2 C 11x C 12 D .x C 3/2 # B.x C 1/.x C 3/ # 3.x C 1/:
Setting x D 0 then yields
12 D 9 # B.1/.3/ # 3.1/
or
B D #2:
To use the method of undetermined coefficients, expand the right-hand side and gather like terms:
3x 2 C 11x C 12 D .1 # B/x 2 C .3 # 4B/x C .6 # 3B/:
Equating x 2 -coefficients on both sides, we find
3D1#B
or
B D #2:
858
TECHNIQUES OF INTEGRATION
CHAPTER 7
4. Find the constants in the partial fraction decomposition
2x C 4
A
Bx C C
D
C 2
x #2
.x # 2/.x 2 C 4/
x C4
SOLUTION
Clearing denominators gives
2x C 4 D A.x 2 C 4/ C .Bx C C /.x # 2/:
Setting x D 2 then yields
4 C 4 D A.4 C 4/ C 0
or
A D 1:
To find B and C , expand the right side, gather like terms, and use the method of undetermined coefficients:
2x C 4 D .B C 1/x 2 C .#2B C C /x C .4 # 2C /:
Equating x 2 -coefficients, we find
or
0DB C1
B D #1;
while equating constants yields
or
4 D 4 # 2C
C D 0:
Thus, A D 1, B D #1, C D 0.
In Exercises 5–8, evaluate using long division first to write f .x/ as the sum of a polynomial and a proper rational function.
Z
x dx
5.
3x # 4
SOLUTION
Long division gives us
x
1
4=3
D C
3x # 4
3
3x # 4
Therefore the integral is
6.
Z
Z
.x 2 C 2/ dx
xC3
SOLUTION
x
dx D
3x # 4
Z
1
4
1
4
#
dx D x # ln j9x # 12j C C
3 9x # 12
3
9
Long division gives us
x2 C 2
11
Dx#3C
:
xC3
xC3
Therefore the integral is
7.
Z
Z
.x 3 C 2x 2 C 1/ dx
xC2
SOLUTION
x2 C 2
dx D
xC3
Z
.x # 3/ dx C 11
Z
dx
x2
D
# 3x C 11 ln jx C 3j C C:
xC3
2
Long division gives us
x 3 C 2x 2 C 1
1
D x2 C
xC2
xC2
Therefore the integral is
8.
Z
.x 3 C 1/ dx
x2 C 1
Z
x 3 C 2x 2 C 1
dx D
xC2
Z
x2 C
1
1
dx D x 3 C ln jx C 2j C C
xC2
3
S E C T I O N 7.5
SOLUTION
The Method of Partial Fractions
Long division gives
x3 C 1
x #1
Dx# 2
x2 C 1
x C1
Therefore the integral is
Z 3
Z
Z
x C1
x#1
1 2
x
1
dx
D
x
#
dx
D
x
#
dx C 2
dx
2
x2 C 1
x2 C 1
x2 C 1
x C1
Z
1
1
2x dx
1
1
1
D x2 #
C 2
dx D x 2 # ln.x 2 C 1/ C tan!1 x C C
2
2
2
2
x2 C 1
x C1
In Exercises 9–44, evaluate the integral.
Z
dx
9.
.x # 2/.x # 4/
SOLUTION
The partial fraction decomposition has the form:
1
A
B
D
C
:
.x # 2/.x # 4/
x#2
x#4
Clearing denominators gives us
1 D A.x # 4/ C B.x # 2/:
Setting x D 2 then yields
1 D A.2 # 4/ C 0
1
AD# ;
2
or
while setting x D 4 yields
1 D 0 C B.4 # 2/
or
BD
1
:
2
The result is:
1
# 12
1
D
C 2 :
.x # 2/.x # 4/
x#2
x#4
Thus,
10.
Z
.x C 3/ dx
xC4
SOLUTION
Z
dx
1
D#
.x # 2/.x # 4/
2
Z
dx
1
C
x#2
2
Z
dx
1
1
D # ln jx # 2j C ln jx # 4j C C:
x#4
2
2
Start with long division:
xC3
1
D1#
xC4
xC4
so that
11.
Z
dx
x.2x C 1/
SOLUTION
Z
xC3
dx D
xC4
Z
1#
1
dx D x # ln jx C 4j C C
xC4
The partial fraction decomposition has the form:
1
A
B
D C
:
x.2x C 1/
x
2x C 1
Clearing denominators gives us
1 D A.2x C 1/ C Bx:
Setting x D 0 then yields
1 D A.1/ C 0
or
A D 1;
859
860
TECHNIQUES OF INTEGRATION
CHAPTER 7
while setting x D # 12 yields
! "
1
1D 0CB #
2
or
B D #2:
The result is:
1
1
#2
D C
:
x.2x C 1/
x
2x C 1
Thus,
Z
dx
D
x.2x C 1/
Z
dx
#
x
Z
2 dx
D ln jxj # ln j2x C 1j C C:
2x C 1
For the integral on the right, we have used the substitution u D 2x C 1, du D 2 dx.
Z
.2x # 1/ dx
12.
x 2 # 5x C 6
SOLUTION
The partial fraction decomposition has the form:
2x # 1
2x # 1
A
B
D
D
C
:
.x # 2/.x # 3/
x#2
x #3
x 2 # 5x C 6
Clearing denominators gives us
2x # 1 D A.x # 3/ C B.x # 2/:
Setting x D 2 then yields
3 D A.#1/ C 0
or
A D #3;
5 D 0 C B.1/
or
B D 5:
while setting x D 3 yields
The result is:
2x # 1
#3
5
D
C
:
x#2
x #3
x 2 # 5x C 6
Thus,
13.
Z
x 2 dx
x2 C 9
Z
.2x # 1/ dx
D #3
x 2 # 5x C 6
Z
dx
C5
x#2
Z
dx
D #3 ln jx # 2j C 5 ln jx # 3j C C:
x#3
SOLUTION
14.
Z
dx
.x # 2/.x # 3/.x C 2/
SOLUTION
Z
x2
dx D
2
x C9
Z
1#
x2
#x $
9
dx D x # 3 tan!1
CC
3
C9
The partial fraction decomposition has the form:
1
A
B
C
D
C
C
:
.x # 2/.x # 3/.x C 2/
x#2
x#3
xC2
Clearing denominators gives us
1 D A.x # 3/.x C 2/ C B.x # 2/.x C 2/ C C.x # 2/.x # 3/:
Setting x D 2 then yields
1 D A.#1/.4/ C 0 C 0
or
1
AD# ;
4
1 D 0 C B.1/.5/ C 0
or
BD
while setting x D 3 yields
1
;
5
The Method of Partial Fractions
S E C T I O N 7.5
and setting x D #2 yields
or
1 D 0 C 0 C C.#4/.#5/
C D
1
:
20
The result is:
1
1
# 14
1
D
C 5 C 20 :
.x # 2/.x # 3/.x C 2/
x#2
x#3
xC2
Thus,
Z
15.
Z
dx
1
D#
.x # 2/.x # 3/.x C 2/
4
Z
dx
1
C
x #2
5
Z
dx
1
C
x#3
20
Z
dx
xC2
1
1
1
D # ln jx # 2j C ln jx # 3j C
ln jx C 2j C C:
4
5
20
.x 2 C 3x # 44/ dx
.x C 3/.x C 5/.3x # 2/
SOLUTION
The partial fraction decomposition has the form:
x 2 C 3x # 44
A
B
C
D
C
C
:
.x C 3/.x C 5/.3x # 2/
xC3
xC5
3x # 2
Clearing denominators gives us
x 2 C 3x # 44 D A.x C 5/.3x # 2/ C B.x C 3/.3x # 2/ C C.x C 3/.x C 5/:
Setting x D #3 then yields
9 # 9 # 44 D A.2/.#11/ C 0 C 0
or
A D 2;
while setting x D #5 yields
25 # 15 # 44 D 0 C B.#2/.#17/ C 0
and setting x D
2
3
or
B D #1;
yields
4
C 2 # 44 D 0 C 0 C C
9
!
11
3
"!
17
3
"
or
C D #2:
The result is:
x 2 C 3x # 44
2
#1
#2
D
C
C
:
.x C 3/.x C 5/.3x # 2/
xC3
xC5
3x # 2
Thus,
Z
.x 2 C 3x # 44/ dx
D2
.x C 3/.x C 5/.3x # 2/
Z
dx
#
xC3
Z
dx
#2
xC5
D 2 ln jx C 3j # ln jx C 5j #
Z
dx
3x # 2
2
ln j3x # 2j C C:
3
To evaluate the last integral, we have made the substitution u D 3x # 2, du D 3 dx.
Z
3 dx
16.
.x C 1/.x 2 C x/
SOLUTION
The partial fraction decomposition has the form:
3
3
3
A
B
C
D
D
D C
C
:
.x C 1/.x/.x C 1/
x
xC1
.x C 1/.x 2 C x/
x.x C 1/2
.x C 1/2
Clearing denominators gives us
3 D A.x C 1/2 C Bx.x C 1/ C C x:
Setting x D 0 then yields
3 D A.1/ C 0 C 0
or
A D 3;
861
862
CHAPTER 7
TECHNIQUES OF INTEGRATION
while setting x D #1 yields
or
3 D 0 C 0 C C.#1/
C D #3:
Now plug in A D 3 and C D #3:
3 D 3.x C 1/2 C Bx.x C 1/ # 3x:
The constant B can be determined by plugging in for x any value other than 0 or #1. Plugging in x D 1 gives us
or
3 D 3.4/ C B.1/.2/ # 3
B D #3:
The result is
3
3
#3
#3
D C
C
:
x
xC1
.x C 1/.x 2 C x/
.x C 1/2
Thus,
17.
Z
Z
3 dx
D3
.x C 1/.x 2 C x/
.x 2 C 11x/ dx
.x # 1/.x C 1/2
SOLUTION
Z
dx
#3
x
Z
dx
#3
xC1
Z
dx
3
D 3 ln jxj # 3 ln jx C 1j C
C C:
xC1
.x C 1/2
The partial fraction decomposition has the form:
x 2 C 11x
A
B
C
D
C
C
:
x#1
xC1
.x # 1/.x C 1/2
.x C 1/2
Clearing denominators gives us
x 2 C 11x D A.x C 1/2 C B.x # 1/.x C 1/ C C.x # 1/:
Setting x D 1 then yields
or
12 D A.4/ C 0 C 0
A D 3;
while setting x D #1 yields
#10 D 0 C 0 C C.#2/
or
C D 5:
Plugging in these values results in
x 2 C 11x D 3.x C 1/2 C B.x # 1/.x C 1/ C 5.x # 1/:
The constant B can be determined by plugging in for x any value other than 1 or #1. If we plug in x D 0, we get
or
0 D 3 C B.#1/.1/ C 5.#1/
B D #2:
The result is
x 2 C 11x
3
#2
5
D
C
C
:
x#1
xC1
.x # 1/.x C 1/2
.x C 1/2
Thus,
18.
Z
Z
.x 2 C 11x/ dx
D3
.x # 1/.x C 1/2
.4x 2 # 21x/ dx
.x # 3/2 .2x C 3/
SOLUTION
Z
dx
#2
x#1
Z
dx
C5
xC1
Z
dx
5
D 3 ln jx # 1j # 2 ln jx C 1j #
C C:
xC1
.x C 1/2
The partial fraction decomposition has the form:
4x 2 # 21x
A
B
C
D
C
C
:
x#3
2x C 3
.x # 3/2 .2x C 3/
.x # 3/2
Clearing denominators gives us
4x 2 # 21x D A.x # 3/.2x C 3/ C B.2x C 3/ C C.x # 3/2 :
Setting x D 3 then yields
#27 D 0 C B.9/ C 0
or
B D #3;
S E C T I O N 7.5
while setting x D # 32 yields
9C
63
D0C0CC
2
!
81
4
"
or
The Method of Partial Fractions
C D 2:
Plugging in these values results in
4x 2 # 21x D A.x # 3/.2x C 3/ # 3.2x C 3/ C 2.x # 3/2 :
Setting x D 0 gives us
0 D A.#3/.3/ # 9 C 18
or
A D 1:
The result is
4x 2 # 21x
1
#3
2
D
C
C
:
x#3
2x C 3
.x # 3/2 .2x C 3/
.x # 3/2
Thus,
19.
Z
Z
.4x 2 # 21x/ dx
D
.x # 3/2 .2x C 3/
dx
.x # 1/2 .x # 2/2
SOLUTION
Z
dx
#3
x#3
Z
dx
C
.x # 3/2
Z
2 dx
3
D ln jx # 3j C
C ln j2x C 3j C C:
2x C 3
x #3
The partial fraction decomposition has the form:
1
A
B
C
D
D
C
C
C
:
x #1
x#2
.x # 1/2 .x # 2/2
.x # 1/2
.x # 2/2
Clearing denominators gives us
1 D A.x # 1/.x # 2/2 C B.x # 2/2 C C.x # 2/.x # 1/2 C D.x # 1/2 :
Setting x D 1 then yields
1 D B.1/
or
B D 1;
1 D D.1/
or
D D 1:
while setting x D 2 yields
Plugging in these values gives us
1 D A.x # 1/.x # 2/2 C .x # 2/2 C C.x # 2/.x # 1/2 C .x # 1/2 :
Setting x D 0 now yields
1 D A.#1/.4/ C 4 C C.#2/.1/ C 1
or
# 4 D #4A # 2C;
while setting x D 3 yields
1 D A.2/.1/ C 1 C C.1/.4/ C 4
or
# 4 D 2A C 4C:
Solving this system of two equations in two unknowns gives A D 2 and C D #2. The result is
1
2
1
#2
1
D
C
C
C
:
x #1
x#2
.x # 1/2 .x # 2/2
.x # 1/2
.x # 2/2
Thus,
Z
20.
Z
dx
D2
.x # 1/2 .x # 2/2
Z
dx
C
x#1
D 2 ln jx # 1j #
.x 2 # 8x/ dx
.x C 1/.x C 4/3
Z
dx
#2
.x # 1/2
Z
dx
C
x#2
Z
dx
.x # 2/2
1
1
# 2 ln jx # 2j #
C C:
x #1
x#2
863
864
CHAPTER 7
SOLUTION
TECHNIQUES OF INTEGRATION
The partial fraction decomposition is
x 2 # 8x
A
B
C
D
D
C
C
C
xC1
xC4
.x C 1/.x C 4/3
.x C 4/2
.x C 4/3
Clearing fractions gives
x 2 # 8x D A.x C 4/3 C B.x C 4/2 .x C 1/ C C.x C 4/.x C 1/ C D.x C 1/
Setting x D #4 gives 48 D #3D so that D D #16. Setting x D #1 gives 9 D 27A so that A D 13 . Thus
x 2 # 8x D
1
.x C 4/3 C B.x C 4/2 .x C 1/ C C.x C 4/.x C 1/ # 16.x C 1/
3
The coefficient of x 3 on the right hand side must be zero; it is 13 C B, so that B D # 13 . Finally, the constant term on the right must
be zero as well; substituting the known values of A, B, and D gives for the constant term
1
1
! 64 # ! 16 C 4C # 16 D 4C
3
3
so that C D 0, and the partial fraction decomposition is
x 2 # 8x
1
1
16
D
#
#
3
3.x C 1/ 3.x C 4/ .x C 4/3
.x C 1/.x C 4/
Thus
Z
21.
Z
x 2 # 8x
1
dx D
3
.x C 1/.x C 4/3
D
8 dx
x.x C 2/3
SOLUTION
Z
1
1
dx #
xC1
3
Z
1
dx # 16
xC4
Z
1
dx
.x C 4/3
ˇ
ˇ
1
1
1 ˇ x C 1 ˇˇ
ln jx C 1j # ln jx C 4j C 8.x C 4/!2 C C D ln ˇˇ
C 8.x C 4/!2 C C
3
3
3
x C 4ˇ
The partial fraction decomposition is
8
A
B
C
D
D C
C
C
3
2
x
xC2
x.x C 2/
.x C 2/
.x C 2/3
Clearing fractions gives
8 D A.x C 2/3 C Bx.x C 2/2 C C x.x C 2/ C Dx
Setting x D 0 gives 8 D 8A so A D 1; setting x D #2 gives 8 D #2D so that D D #4; the result is
8 D .x C 2/3 C Bx.x C 2/2 C C x.x C 2/ # 4x
The coefficient of x 3 on the right-hand side must be zero, since it is zero on the left. We compute it to be 1 C B, so that B D #1.
Finally, we look at the coefficient of x 2 on the right-hand side; it must be zero as well. We compute it to be
3!2#4CC D C C2
so that C D #2 and the partial fraction decomposition is
8
1
1
2
4
D #
#
#
3
2
x x C 2 .x C 2/
x.x C 2/
.x C 2/3
and
Z
22.
Z
8
dx D
x.x C 2/3
Z
1
1
dx #
dx # 2
x
xC2
Z
.x C 2/!2 dx # 4
Z
.x C 2/!3 dx
ˇ
ˇ
D ln jxj # ln jx C 2j C 2.x C 2/!1 C 2.x C 2/!2 C C D ln ˇˇ
x 2 dx
x2 C 3
ˇ
x ˇˇ
2
2
C
C
CC
x C 2ˇ x C 2
.x C 2/2
SOLUTION
Z
x2
dx D
2
x C3
Z
1#
3
dx D
x2 C 3
Z
1 dx # 3
Z
p
1
dx D x # 3 tan!1
2
x C3
!
x
p
3
"
CC
The Method of Partial Fractions
S E C T I O N 7.5
Z
dx
2x 2 # 3
SOLUTION The partial fraction decomposition has the form
23.
1
1
A
B
D p
p p
p D p
p Cp
p :
2x 2 # 3
. 2x # 3/. 2x C 3/
2x # 3
2x C 3
Clearing denominators, we get
1DA
Setting x D
p p
3= 2 then yields
1DA
p p
while setting x D # 3= 2 yields
The result is
#p
#p
p $
p $
2x C 3 C B
2x # 3 :
#p
3C
p $
3 C0
1
AD p ;
2 3
or
# p
p $
1 D 0CB # 3# 3
#1
BD p :
2 3
or
p
p
1
1=2 3
1=2 3
D
p
p
#
p
p :
2x 2 # 3
2x # 3
2x C 3
Thus,
Z
Z
Z
dx
1
dx
1
dx
D
p
p
p
#
p
p
p :
2x 2 # 3
2 3
2x # 3 2 3
2x C 3
p
p
p
p
p
p
For the first integral, let u D 2x # 3, du D 2 dx, and for the second, let w D 2x C 3, dw D 2 dx. Then we have
Z
Z
Z
ˇp
ˇp
p ˇˇ
p ˇˇ
dx
1
du
1
dw
1
1
ˇ
ˇ
D
p
p
#
p
p
D
p
ln
2x
#
3
#
p
ln
2x
C
3ˇ C C:
ˇ
ˇ
ˇ
u
w
2x 2 # 3
2 3. 2/
2 3. 2/
2 6
2 6
Z
dx
24.
.x # 4/2 .x # 1/
SOLUTION
The partial fraction decomposition has the form:
1
A
B
C
D
C
C
:
x#4
.x # 1/
.x # 4/2 .x # 1/
.x # 4/2
Clearing denominators, we get
1 D A.x # 4/.x # 1/ C B.x # 1/ C C.x # 4/2 :
Setting x D 1 then yields
1 D 0 C 0 C C.9/
or
C D
1
;
9
1 D 0 C B.3/ C 0
or
BD
1
:
3
while setting x D 4 yields
Plugging in B D
1
3
and C D 19 , and setting x D 5, we find
1
1
1 D A.1/.4/ C .4/ C .1/
3
9
or
1
AD# :
9
The result is
1
.x # 4/2 .x # 1/
D
1
1
# 19
3
9
C
C
:
x#4
x #1
.x # 4/2
Thus,
Z
1
dx
D#
9
.x # 4/2 .x # 1/
Z
dx
1
C
x#4
3
Z
1
dx
C
9
.x # 4/2
Z
dx
x#1
1
1
1
C ln jx # 1j C C:
D # ln jx # 4j #
9
3.x # 4/
9
865
866
25.
TECHNIQUES OF INTEGRATION
CHAPTER 7
Z
4x 2 # 20
dx
.2x C 5/3
SOLUTION
The partial fraction decomposition is
4x 2 # 20
A
B
C
D
C
C
2x C 5
.2x C 5/3
.2x C 5/2
.2x C 5/3
Clearing fractions gives
4x 2 # 20 D A.2x C 5/2 C B.2x C 5/ C C
Setting x D #5=2 gives 5 D C so that C D 5. The coefficient of x 2 on the left-hand side is 4, and on the right-hand side is 4A, so
that A D 1 and we have
4x 2 # 20 D .2x C 5/2 C B.2x C 5/ C 5
Considering the constant terms now gives #20 D 25 C 5B C 5 so that B D #10. Thus
Z
Z
Z
Z
4x 2 # 20
1
1
1
D
dx
#
10
dx
C
5
dx
2x C 5
.2x C 5/3
.2x C 5/2
.2x C 5/3
26.
Z
D
x 2 .x
SOLUTION
3x C 6
dx
# 1/.x # 3/
1
5
5
ln j2x C 5j C
#
CC
2
2x C 5 4.2x C 5/2
The partial fraction decomposition has the form:
3x C 6
A
B
C
D
D C 2 C
C
:
x
x #1
x #3
x 2 .x # 1/.x # 3/
x
Clearing denominators gives us
3x C 6 D Ax.x # 1/.x # 3/ C B.x # 1/.x # 3/ C C x 2 .x # 3/ C Dx 2 .x # 1/:
Setting x D 0, then yields
or
6 D 0 C B.#1/.#3/ C 0 C 0
B D 2;
while setting x D 1 yields
9 D 0 C 0 C C.1/.#2/ C 0
or
9
C D# ;
2
15 D 0 C 0 C 0 C D.9/.2/
or
DD
and setting x D 3 yields
5
:
6
In order to find A, let’s look at the x 3 -coefficient on the right-hand side (which must equal 0, since there’s no x 3 term on the left):
0DACC CD DA#
9
5
C ;
2
6
so
AD
11
:
3
The result is
x 2 .x
11
5
# 92
3x C 6
2
D 3 C 2 C
C 6 :
x
x #1
x #3
# 1/.x # 3/
x
Thus,
Z
27.
Z
.3x C 6/ dx
11
D
2
3
x .x # 1/.x # 3/
D
dx
x.x # 1/3
Z
dx
C2
x
Z
dx 9
#
2
x2
Z
dx
5
C
x#1
6
Z
dx
x#3
2 9
5
11
ln jxj # # ln jx # 1j C ln jx # 3j C C:
3
x 2
6
The Method of Partial Fractions
S E C T I O N 7.5
SOLUTION
The partial fraction decomposition has the form:
1
A
B
C
D
D C
C
C
:
x
x#1
x.x # 1/3
.x # 1/2
.x # 1/3
Clearing denominators, we get
1 D A.x # 1/3 C Bx.x # 1/2 C C x.x # 1/ C Dx:
Setting x D 0 then yields
1 D A.#1/ C 0 C 0 C 0
or
A D #1;
1 D 0 C 0 C 0 C D.1/
or
D D 1:
while setting x D 1 yields
Plugging in A D #1 and D D 1 gives us
1 D #.x # 1/3 C Bx.x # 1/2 C C x.x # 1/ C x:
Now, setting x D 2 yields
1 D #1 C 2B C 2C C 2
or
2B C 2C D 0;
and setting x D 3 yields
or
1 D #8 C 12B C 6C C 3
2B C C D 1:
Solving these two equations in two unknowns, we find B D 1 and C D #1. The result is
1
#1
1
#1
1
D
C
C
C
:
x
x#1
x.x # 1/3
.x # 1/2
.x # 1/3
Thus,
Z
dx
D#
x.x # 1/3
Z
dx
C
x
Z
dx
#
x #1
D # ln jxj C ln jx # 1j C
28.
Z
.3x 2 # 2/ dx
x#4
SOLUTION
Z
dx
C
.x # 1/2
Z
dx
.x # 1/3
1
1
#
C C:
x # 1 2.x # 1/2
First we use long division to write
3x 2 # 2
46
D 3x C 12 C
:
x#4
x#4
Then the integral becomes
Z
Z
Z
.3x 2 # 2/ dx
dx
3
D .3x C 12/ dx C 46
D x 2 C 12x C 46 ln jx # 4j C C:
x#4
x #4
2
Z
.x 2 # x C 1/ dx
29.
x2 C x
SOLUTION
First use long division to write
x2 # x C 1
#2x C 1
#2x C 1
D1C 2
D1C
:
x.x C 1/
x2 C x
x Cx
The partial fraction decomposition of the term on the right has the form:
#2x C 1
A
B
D C
:
x.x C 1/
x
xC1
Clearing denominators gives us
#2x C 1 D A.x C 1/ C Bx:
Setting x D 0 then yields
1 D A.1/ C 0
or
A D 1;
867
868
TECHNIQUES OF INTEGRATION
CHAPTER 7
while setting x D #1 yields
or
3 D 0 C B.#1/
B D #3:
The result is
#2x C 1
1
#3
D C
:
x.x C 1/
x
xC1
Thus,
30.
Z
dx
x.x 2 C 1/
SOLUTION
Z
.x 2 # x C 1/ dx
D
x2 C x
Z
dx C
Z
dx
#3
x
Z
dx
D x C ln jxj # 3 ln jx C 1j C C:
xC1
The partial fraction decomposition has the form:
1
A
Bx C C
D C 2
:
x
x.x 2 C 1/
x C1
Clearing denominators, we get
1 D A.x 2 C 1/ C .Bx C C /x:
Setting x D 0 then yields
or
1 D A.1/ C 0
A D 1:
This gives us
1 D x 2 C 1 C Bx 2 C C x D .B C 1/x 2 C C x C 1:
Equating x 2 -coefficients, we find
or
B C1D0
B D #1I
while equating x-coefficients yields C D 0. The result is
1
1
#x
D C 2
:
x
x.x 2 C 1/
x C1
Thus,
Z
dx
D
x.x 2 C 1/
Z
dx
#
x
Z
x dx
:
x2 C 1
For the integral on the right, use the substitution u D x 2 C 1, du D 2x dx. Then we have
Z
Z
Z
dx
dx 1
du
1
D
#
D ln jxj # ln jx 2 C 1j C C:
x
2
u
2
x.x 2 C 1/
Z
2
.3x # 4x C 5/ dx
31.
.x # 1/.x 2 C 1/
SOLUTION
The partial fraction decomposition has the form:
3x 2 # 4x C 5
A
Bx C C
D
C 2
:
x#1
.x # 1/.x 2 C 1/
x C1
Clearing denominators, we get
3x 2 # 4x C 5 D A.x 2 C 1/ C .Bx C C /.x # 1/:
Setting x D 1 then yields
3 # 4 C 5 D A.2/ C 0
or
A D 2:
This gives us
3x 2 # 4x C 5 D 2.x 2 C 1/ C .Bx C C /.x # 1/ D .B C 2/x 2 C .C # B/x C .2 # C /:
The Method of Partial Fractions
S E C T I O N 7.5
Equating x 2 -coefficients, we find
3DB C2
or
B D 1I
while equating constant coefficients yields
5D2#C
or
C D #3:
The result is
3x 2 # 4x C 5
2
x#3
D
C 2
:
2
x
#
1
.x # 1/.x C 1/
x C1
Thus,
Z
.3x 2 # 4x C 5/ dx
D2
.x # 1/.x 2 C 1/
Z
dx
C
x#1
Z
.x # 3/ dx
D2
x2 C 1
Z
dx
C
x#1
Z
x dx
#3
x2 C 1
Z
dx
:
x2 C 1
For the second integral, use the substitution u D x 2 C 1, du D 2x dx. The final answer is
Z
.3x 2 # 4x C 5/ dx
1
D 2 ln jx # 1j C ln jx 2 C 1j # 3 tan!1 x C C:
2
2
.x # 1/.x C 1/
Z
2
x
32.
dx
.x C 1/.x 2 C 1/
SOLUTION
The partial fraction decomposition has the form
x2
A
Bx C C
D
C 2
:
xC1
.x C 1/.x 2 C 1/
x C1
Clearing denominators, we get
x 2 D A.x 2 C 1/ C .Bx C C /.x C 1/:
Setting x D #1 then yields
or
1 D A.2/ C 0
AD
1
:
2
This gives us
x2 D
!
"
!
"
1 2 1
1
1
x C C Bx 2 C Bx C C x C C D B C
x 2 C .B C C /x C C C
:
2
2
2
2
Equating x 2 -coefficients, we find
1DBC
1
2
or
BD
1
;
2
while equating constant coefficients yields
0DC C
1
2
or
1
C D# :
2
The result is
1
1
x# 1
x2
D 2 C 22 2 :
2
xC1
.x C 1/.x C 1/
x C1
Thus,
Z
x 2 dx
1
D
2
.x C 1/.x 2 C 1/
D
Z
dx
1
C
xC1
2
Z
.x # 1/ dx
1
D
2
x2 C 1
Z
dx
1
C
xC1
2
1
1
1
ln jx C 1j C ln jx 2 C 1j # tan!1 x C C:
2
4
2
Here we used u D x 2 C 1, du D 2x dx for the second integral.
Z
dx
33.
x.x 2 C 25/
Z
x dx
1
#
x2 C 1 2
Z
dx
x2 C 1
869
870
CHAPTER 7
SOLUTION
TECHNIQUES OF INTEGRATION
The partial fraction decomposition has the form:
1
A
Bx C C
D C 2
:
x
x.x 2 C 25/
x C 25
Clearing denominators, we get
1 D A.x 2 C 25/ C .Bx C C /x:
Setting x D 0 then yields
or
1 D A.25/ C 0
AD
1
:
25
This gives us
1D
!
"
1 2
1
x C 1 C Bx 2 C C x D B C
x 2 C C x C 1:
25
25
Equating x 2 -coefficients, we find
0DBC
1
25
or
BD#
1
;
25
while equating x-coefficients yields C D 0. The result is
1
1
# 25
x
1
25
D
C
:
x
x.x 2 C 25/
x 2 C 25
Thus,
Z
dx
1
D
25
x.x 2 C 25/
Z
dx
1
#
x
25
Z
x dx
:
x 2 C 25
For the integral on the right, use u D x 2 C 25, du D 2x dx. Then we have
Z
dx
1
1
D
ln jxj #
ln jx 2 C 25j C C:
x.x 2 C 25/
25
50
Z
dx
34.
x 2 .x 2 C 25/
SOLUTION
The partial fraction decomposition has the form:
1
A
B
Cx C D
D C 2 C 2
:
x
x 2 .x 2 C 25/
x
x C 25
Clearing denominators, we get
1 D Ax.x 2 C 25/ C B.x 2 C 25/ C .C x C D/x 2 :
Setting x D 0 then yields
or
1 D 0 C B.25/ C 0
BD
1
:
25
This gives us
1 D Ax 3 C 25Ax C
!
"
1 2
1
x C 1 C C x 3 C Dx 2 D .A C C /x 3 C D C
x 2 C 25Ax C 1:
25
25
Equating x-coefficients yields
0 D 25A
or
A D 0;
while equating x 3 -coefficients yields
0DACC D 0CC
or
C D 0;
and equating x 2 -coefficients yields
0DDC
1
25
or
DD
#1
:
25
The Method of Partial Fractions
S E C T I O N 7.5
The result is
1
!1
1
25
25
D
C
:
x 2 .x 2 C 25/
x2
x 2 C 25
Thus,
35.
Z
.6x 2 C 2/ dx
x 2 C 2x # 3
Z
dx
1
D
25
x 2 .x 2 C 25/
Z
dx
1
#
25
x2
Z
# $
dx
1
1
!1 x
D
#
#
tan
C C:
25x
125
5
x 2 C 25
Long division gives
SOLUTION
6x 2 C 2
12x # 20
12x # 20
D6# 2
D6#
.x C 3/.x # 1/
x 2 C 2x # 3
x C 2x # 3
The partial fraction decomposition of the second term is
12x # 20
A
B
D
C
.x C 3/.x # 1/
xC3
x #1
Clear fractions to get
12x # 20 D A.x # 1/ C B.x C 3/
Set x D 1 to get #8 D 4B so that B D #2. Set x D #3 to get #56 D #4A so that A D 14, and we have
Z
Z
Z
Z
Z
6x 2 C 2
14
2
1
1
D
6
#
C
dx
D
6
dx
#
14
dx
C
2
dx
xC3
x#1
xC3
x #1
x 2 C 2x # 3
36.
Z
D 6x # 14 ln jx C 3j C 2 ln jx # 1j C C
6x 2 C 7x # 6
dx
.x 2 # 4/.x C 2/
The partial fraction decomposition has the form:
SOLUTION
6x 2 C 7x # 6
6x 2 C 7x # 6
A
B
C
D
D
C
C
:
2
.x # 2/.x C 2/.x C 2/
x #2
xC2
.x # 4/.x C 2/
.x C 2/2
Clearing denominators, we get
6x 2 C 7x # 6 D A.x C 2/2 C B.x # 2/.x C 2/ C C.x # 2/:
Setting x D 2 then yields
24 C 14 # 6 D A.16/ C 0 C 0
or
A D 2;
while setting x D #2 yields
24 # 14 # 6 D 0 C 0 C C.#4/
or
C D #1:
This gives us
6x 2 C 7x # 6 D 2.x C 2/2 C B.x # 2/.x C 2/ # .x # 2/:
Now, setting x D 1 yields
6 C 7 # 6 D 2.9/ C B.#1/.3/ # .#1/
or
B D 4:
The result is
2
6x 2 C 7x # 6
4
#1
D
:
C
C
2
x#2
xC2
.x # 4/.x C 2/
.x C 2/2
Thus,
Z
.6x 2 C 7x # 6/ dx
D2
.x 2 # 4/.x C 2/
Z
dx
C4
x#2
Z
dx
#
xC2
Z
dx
1
D 2 ln jx # 2j C 4 ln jx C 2j C
C C:
xC2
.x C 2/2
871
872
37.
TECHNIQUES OF INTEGRATION
CHAPTER 7
Z
10 dx
.x # 1/2 .x 2 C 9/
SOLUTION
The partial fraction decomposition has the form:
10
.x
# 1/2 .x 2
C 9/
D
A
B
Cx C D
C
C 2
:
2
x#1
.x # 1/
x C9
Clearing denominators, we get
10 D A.x # 1/.x 2 C 9/ C B.x 2 C 9/ C .C x C D/.x # 1/2 :
Setting x D 1 then yields
or
10 D 0 C B.10/ C 0
B D 1:
Expanding the right-hand side, we have
10 D .A C C /x 3 C .1 # A # 2C C D/x 2 C .9A C C # 2D/x C .9 # 9A C D/:
Equating coefficients of like powers of x then yields
ACC D 0
1 # A # 2C C D D 0
9A C C # 2D D 0
9 # 9A C D D 10
From the first equation, we have C D #A, and from the fourth equation we have D D 1 C 9A. Substituting these into the second
equation, we get
Finally, C D
1
5
1
AD# :
5
or
1 # A # 2.#A/ C .1 C 9A/ D 0
and D D # 45 . The result is
1
4
# 15
10
1
5x # 5
D
C
C
:
x#1
.x # 1/2 .x 2 C 9/
.x # 1/2
x2 C 9
Thus,
Z
38.
Z
10 dx
1
D#
5
.x # 1/2 .x 2 C 9/
dx
C
x #1
Z
dx
1
C
5
.x # 1/2
Z
x dx
4
#
x2 C 9 5
Z
dx
x2 C 9
#x$
1
1
1
4
D # ln jx # 1j #
C
ln jx 2 C 9j #
tan!1
C C:
5
x#1
10
15
3
10 dx
.x C 1/.x 2 C 9/2
SOLUTION
Z
The partial fraction decomposition has the form:
10
A
Bx C C
Dx C E
D
C 2
C 2
:
xC1
.x C 1/.x 2 C 9/2
x C9
.x C 9/2
Clearing denominators gives us
10 D A.x 2 C 9/2 C .Bx C C /.x C 1/.x 2 C 9/ C .Dx C E/.x C 1/:
Setting x D #1 then yields
10 D A.100/ C 0 C 0
or
AD
1
:
10
Expanding the right-hand side, we find
!
!
!
"
"
"
18
81
1
10 D B C
x 4 C .B C C /x 3 C 9B C C C D C
x 2 .9B C 9C C D C E/x C 9C C E C
:
10
10
10
Equating x 4 -coefficients yields
BC
1
D0
10
or
BD#
1
;
10
The Method of Partial Fractions
S E C T I O N 7.5
while equating x 3 -coefficients yields
#
1
CC D0
10
or
C D
1
;
10
or
D D #1:
and equating x 2 -coefficients yields
#
9
1
18
C
CDC
D0
10
10
10
Finally, equating constant coefficients, we find
10 D
9
81
CE C
10
10
or
E D 1:
The result is
1
1
1
# 10
x C 10
10
#x C 1
10
D
C
C 2
:
2
2
2
x
C
1
.x C 1/.x C 9/
x C9
.x C 9/2
Thus,
Z
10 dx
1
D
10
.x C 1/.x 2 C 9/2
Z
dx
1
#
x C 1 10
Z
x dx
1
C
10
x2 C 9
Z
dx
#
x2 C 9
Z
x dx
C
.x 2 C 9/2
Z
dx
:
.x 2 C 9/2
For the second and fourth integrals, use the substitution u D x 2 C 9, du D 2x dx. Then we have
Z
Z
# $
10 dx
1
1
1
1
dx
2
!1 x
D
ln
jx
C
1j
#
ln
jx
C
9j
C
tan
C
C
:
2
2
2
2
10
20
30
3
.x C 1/.x C 9/
2.x C 9/
.x C 9/2
For the last integral, use the trigonometric substitution
x D 3 tan !;
dx D 3 sec2 ! d!;
x 2 C 9 D 9 tan2 ! C 9 D 9 sec2 !:
Then,
Z
dx
D
.x 2 C 9/2
Z
3 sec2 ! d!
1
D
27
.9 sec2 !/2
Now we construct a right triangle with tan ! D
Z
d!
1
D
27
sec2 !
Z
%
&
1 1
1
cos ! d! D
! C sin ! cos ! C C:
27 2
2
2
x
3:
x2 + 9
x
3
p
p
From this we see that sin ! D x= x 2 C 9 and cos ! D 3= x 2 C 9. Thus
!
"!
"
Z
# $
# $
dx
1
1
x
3
1
x
!1 x
!1 x
D
tan
C
p
p
C
C
D
tan
C
C C:
2 C 9/
2
2
54
3
54
54
3
.x 2 C 9/2
18.x
x C9
x C9
Collecting all the terms, we obtain
Z
# $
10 dx
1
1
1
1
2
!1 x
D
ln
jx
C
1j
#
ln
jx
C
9j
C
tan
C
10
20
30
3
.x C 1/.x 2 C 9/2
2.x 2 C 9/
#x $
1
x
C
tan!1
C
CC
54
3
18.x 2 C 9/
#x $
1
1
7
xC9
D
ln jx C 1j #
ln jx 2 C 9j C
tan!1
C
C C:
10
20
135
3
18.x 2 C 9/
Z
dx
39.
2
x.x C 8/2
SOLUTION
The partial fraction decomposition has the form:
A
1
Bx C C
Dx C E
D C 2
:
C 2
x
x.x 2 C 8/2
x C8
.x C 8/2
Clearing denominators, we get
1 D A.x 2 C 8/2 C .Bx C C /x.x 2 C 8/ C .Dx C E/x:
873
874
CHAPTER 7
TECHNIQUES OF INTEGRATION
Expanding the right-hand side gives us
1 D .A C B/x 4 C C x 3 C .16A C 8B C D/x 2 C .8C C E/x C 64A:
Equating coefficients of like powers of x yields
ACB D0
C D0
16A C 8B C D D 0
8C C E D 0
64A D 1
The solution to this system of equations is
AD
1
;
64
BD#
1
;
64
1
DD# ;
8
C D 0;
E D 0:
Therefore
x.x 2
1
#1x
#1x
1
D 64 C 264 C 2 8 2 ;
2
x
C 8/
x C8
.x C 8/
and
Z
dx
1
D
64
x.x 2 C 8/2
Z
dx
1
#
x
64
Z
x dx
1
#
x2 C 8 8
Z
x dx
:
.x 2 C 8/2
For the second and third integrals, use the substitution u D x 2 C 8, du D 2x dx. Then we have
Z
dx
1
1
1
D
ln jxj #
ln jx 2 C 8j C
C C:
64
128
x.x 2 C 8/2
16.x 2 C 8/
Z
100x dx
40.
.x # 3/.x 2 C 1/2
SOLUTION
The partial fraction decomposition has the form:
100x
A
Bx C C
Dx C E
D
C 2
C 2
:
x#3
.x # 3/.x 2 C 1/2
x C1
.x C 1/2
Clearing denominators, we get
100x D A.x 2 C 1/2 C .Bx C C /.x # 3/.x 2 C 1/ C .Dx C E/.x # 3/:
Setting x D 3 then yields
300 D A.100/ C 0 C 0
or
A D 3:
Expanding the right-hand side, we find
100x D .B C 3/x 4 C .C # 3B/x 3 C .B # 3C C D C 6/x 2 C .C # 3B # 3D C E/x C .3 # 3C # 3E/:
Equating coefficients of like powers of x then yields
B C3D0
C # 3B D 0
B # 3C C D C 6 D 0
C # 3B # 3D C E D 100
3 # 3C # 3E D 0
The solution to this system of equations is
B D #3;
C D #9;
D D #30;
E D 10:
Therefore
100x
#3x # 9
#30x C 10
3
C 2
C
D
;
x#3
.x # 3/.x 2 C 1/2
x C1
.x 2 C 1/2
S E C T I O N 7.5
The Method of Partial Fractions
and
Z
Z
.#3x # 9/ dx
.#30x C 10/ dx
C
2
x C1
.x 2 C 1/2
Z
Z
Z
Z
Z
dx
x dx
dx
x dx
dx
D3
#3
#
9
#
30
C
10
:
2
2
2
2
2
x #3
x C1
x C1
.x C 1/
.x C 1/2
100x dx
D3
.x # 3/.x 2 C 1/2
Z
dx
C
x #3
Z
For the second and fourth integrals, use the substitution u D x 2 C 1, du D 2x dx. Then we have
Z
Z
100x dx
3
15
dx
2
!1
D
3
ln
jx
#
3j
#
ln
jx
C
1j
#
9
tan
x
C
C
10
:
2
.x # 3/.x 2 C 1/2
x2 C 1
.x 2 C 1/2
For the last integral, use the trigonometric substitution x D tan !, dx D sec2 ! d!. Then x 2 C 1 D tan2 ! C 1 D sec2 !, and
Z
Z
Z
dx
sec2 ! d!
1
1
D
D
cos2 ! D ! C sin ! cos ! C C:
2
2
.x 2 C 1/2
sec4 !
We construct the following right triangle with tan ! D x:
1 + x2
x
1
p
p
From this we see that sin ! D x= 1 C x 2 and cos ! D 1= 1 C x 2 . Thus
!
"!
"
Z
dx
1
1
x
1
1
x
!1
D
tan
x
C
p
p
C C D tan!1 x C
C C:
2
2
2
2
2
2
2
2
.x C 1/
2.x C 1/
1Cx
1Cx
Collecting all the terms, we obtain
!
"
Z
100x dx
3
15
1
x
2
!1
!1
D
3
ln
jx
#
3j
#
ln
jx
C
1j
#
9
tan
x
C
C
10
tan
x
C
CC
2
2
.x # 3/.x 2 C 1/2
x2 C 1
2.x 2 C 1/
41.
Z
D 3 ln jx # 3j #
.x
dx
C 4x C 10/
3
5x C 15
ln jx 2 C 1j # 4 tan!1 x C 2
C C:
2
x C1
C 2/.x 2
SOLUTION
The partial fraction decomposition has the form:
1
A
Bx C C
D
C 2
:
xC2
.x C 2/.x 2 C 4x C 10/
x C 4x C 10
Clearing denominators, we get
1 D A.x 2 C 4x C 10/ C .Bx C C /.x C 2/:
Setting x D #2 then yields
or
1 D A.6/ C 0
AD
1
:
6
Expanding the right-hand side gives us
1D
!
"
!
"
!
"
1
2
5
C B x2 C
C 2B C C x C
C 2C :
6
3
3
Equating x 2 -coefficients yields
0D
1
CB
6
or
1
BD# ;
6
1D
5
C 2C
3
or
1
C D# :
3
while equating constant coefficients yields
The result is
.x
C 2/.x 2
1
# 1 x # 13
1
D 6 C 2 6
:
xC2
x C 4x C 10
C 4x C 10/
875
876
CHAPTER 7
TECHNIQUES OF INTEGRATION
Thus,
Z
dx
1
D
6
.x C 2/.x 2 C 4x C 10/
Z
dx
1
#
xC2 6
Z
.x C 2/ dx
:
x 2 C 4x C 10
For the second integral, let u D x 2 C 4x C 10. Then du D .2x C 4/ dx, and
Z
Z
dx
1
1
.2x C 4/ dx
D
ln
jx
C
2j
#
6
12
.x C 2/.x 2 C 4x C 10/
x 2 C 4x C 10
42.
Z
D
1
1
ln jx C 2j #
ln jx 2 C 4x C 10j C C:
6
12
9 dx
.x C 1/.x 2 # 2x C 6/
SOLUTION
The partial fraction decomposition has the form:
9
.x
C 1/.x 2
# 2x C 6/
A
Bx C C
C 2
:
xC1
x # 2x C 6
D
Clearing denominators gives us
9 D A.x 2 # 2x C 6/ C .Bx C C /.x C 1/:
Setting x D #1 then yields
or
9 D A.9/ C 0
A D 1:
Expanding the right-hand side gives us
9 D .1 C B/x 2 C .#2 C B C C /x C .6 C C /:
Equating x 2 -coefficients yields
0D1CB
or
B D #1;
while equating constant coefficients yields
9D6CC
or
C D 3:
The result is
9
1
#x C 3
D
C 2
:
xC1
.x C 1/.x 2 # 2x C 6/
x # 2x C 6
Thus,
Z
9 dx
D
.x C 1/.x 2 # 2x C 6/
Z
dx
C
xC1
Z
.#x C 3/ dx
:
x 2 # 2x C 6
To evaluate the integral on the right, we first write
Z
Z
Z
Z
.#x C 3/ dx
.x # 1 # 2/ dx
.x # 1/ dx
dx
D
#
D
#
C
2
:
2
2
2
2
x # 2x C 6
x # 2x C 6
x # 2x C 6
x # 2x C 6
For the first integral, use the substitution u D x 2 # 2x C 6, du D .2x # 2/ dx. Then
Z
Z
.x # 1/ dx
1
.2x # 2/ dx
1
#
D
#
D # ln jx 2 # 2x C 6j C C:
2
2
2
2
x # 2x C 6
x # 2x C 6
For the second integral, we first complete the square:
Z
Z
Z
dx
dx
dx
2
D
2
D
2
:
x 2 # 2x C 6
.x 2 # 2x C 1/ C 5
.x # 1/2 C 5
Now let u D x # 1, du D dx. Then
!
"
"
"
!
!
Z
Z
dx
du
1
2
u
!1
!1 x # 1
2
D
2
D
2
p
tan
p
C
C
D
p
tan
p
C C:
.x # 1/2 C 5
u2 C 5
5
5
5
5
Collecting all the terms, we have
!
"
Z
1
2
9 dx
2
!1 x # 1
D
ln
jx
C
1j
#
ln
jx
#
2x
C
6j
C
p
tan
p
C C:
2
.x C 1/.x 2 # 2x C 6/
5
5
S E C T I O N 7.5
43.
Z
The Method of Partial Fractions
877
25 dx
x.x 2 C 2x C 5/2
SOLUTION
The partial fraction decomposition has the form
x.x 2
25
A
Bx C C
Dx C E
D C 2
C 2
:
2
x
C 2x C 5/
x C 2x C 5
.x C 2x C 5/2
Clearing denominators yields:
25 D A.x 2 C 2x C 5/2 C x.Bx C C /.x 2 C 2x C 5/ C x.Dx C E/
D .Ax 4 C 4Ax 3 C 14Ax 2 C 20Ax C 25A/ C .Bx 4 C C x 3 C 2Bx 3 C 2C x 2 C 5Bx 2 C 5C x/ C Dx 2 C Ex:
Equating constant terms yields
or
25A D 25
A D 1;
while equating x 4 -coefficients yields
ACB D0
or
B D #A D #1:
Equating x 3 -coefficients yields
4A C C C 2B D 0
or
C D #2;
and equating x 2 -coefficients yields
or
14A C 2C C 5B C D D 0
D D #5:
Finally, equating x-coefficients yields
20A C 5C C E D 0
or
E D #10:
Thus,
Z
"
1
xC2
xC2
# 2
#5 2
dx
x x C 2x C 5
.x C 2x C 5/2
Z
Z
xC2
xC2
D ln jxj #
dx
#
5
dx:
x 2 C 2x C 5
.x 2 C 2x C 5/2
25 dx
D
x.x 2 C 2x C 5/2
Z !
The two integrals on the right both require the substitution u D x C 1, so that x 2 C 2x C 5 D .x C 1/2 C 4 D u2 C 4 and
du D dx. This means:
Z
Z
Z
25 dx
uC1
uC1
D
ln
jxj
#
du
#
5
du
x.x 2 C 2x C 5/2
u2 C 4
.u2 C 4/2
Z
Z
Z
Z
u
1
u
1
D ln jxj #
du
#
du
#
5
du
#
5
du:
2
2
2
2
2
u C4
u C4
.u C 4/
.u C 4/2
For the first and third integrals, we make the substitution w D u2 C 4, dw D 2u du. Then we have
Z
Z
# $
25 dx
1
1
5
du
2
!1 u
D ln jxj # ln ju C 4j # tan
C
#5
2
2
2
x.x 2 C 2x C 5/2
2.u2 C 4/
.u2 C 4/2
!
"
Z
1
1
5
du
2
!1 x C 1
D ln jxj # ln jx C 2x C 5j # tan
C
#
5
:
2
2
2
2.x 2 C 2x C 5/
.u2 C 4/2
For the remaining integral, we use the trigonometric substitution 2 tan w D u, so that u2 C 4 D 4 tan2 w C 4 D 4 sec2 w and
du D 2 sec2 w dw. This means
Z
Z
Z
1
1
1
1
2
du
D
sec
w
dw
D
cos2 w dw
8
8
.u2 C 4/2
sec4 w
!
"
!
"
1 1
w
1
w
D
sin 2w C
CC D
sin w cos w C
CC
8 4
2
16
16
#u$
#u$
1
u
2
1
1 u
1
D
C
tan!1
CC D
C
tan!1
CC
p
p
2
16 u2 C 4 u2 C 4
16
2
8u C4
16
2
"
!
1
xC1
1
!1 x C 1
D
C
tan
:
8 x 2 C 2x C 5
16
2
878
CHAPTER 7
TECHNIQUES OF INTEGRATION
Hence, the integral is
Z
25 dx
1
1
D ln jxj # ln jx 2 C 2x C 5j # tan!1
2
2
x.x 2 C 2x C 5/2
!
xC1
2
"
!
"
5
5
xC1
5
!1 x C 1
#
#
tan
2
2.x 2 C 2x C 5/ 8 x 2 C 2x C 5 16
!
"
15 # 5x
13
xC1
1
D ln jxj C
#
tan!1
# ln jx 2 C 2x C 5j C C:
2
2
8.x 2 C 2x C 5/ 16
C
44.
Z
.x 2 C 3/ dx
.x 2 C 2x C 3/2
SOLUTION
The partial fraction decomposition has the form:
.x 2
x2 C 3
Ax C B
Cx C D
D 2
C 2
:
C 2x C 3/2
x C 2x C 3
.x C 2x C 3/2
Clearing denominators gives us
x 2 C 3 D .Ax C B/.x 2 C 2x C 3/ C C x C D:
Expanding the right-hand side, we get
x 2 C 3 D Ax 3 C .2A C B/x 2 C .3A C 2B C C /x C .3B C D/:
Equating coefficients of like powers of x then yields
AD0
2A C B D 1
3A C 2B C C D 0
3B C D D 3
The solution to this system of equations is
A D 0;
B D 1;
C D #2;
D D 0:
Therefore
x2 C 3
1
#2x
D 2
C 2
;
.x 2 C 2x C 3/2
x C 2x C 3
.x C 2x C 3/2
and
Z
.x 2 C 3/ dx
D
.x 2 C 2x C 3/2
Z
dx
#
x 2 C 2x C 3
Z
2x dx
:
.x 2 C 2x C 3/2
The first integral can be evaluated by completing the square:
Z
Z
Z
dx
dx
dx
D
D
:
2
2
x C 2x C 3
x C 2x C 1 C 2
.x C 1/2 C 2
Now use the substitution u D x C 1, du D dx. Then
!
"
Z
Z
dx
du
1
!1 x C 1
D
D p tan
p
C C:
x 2 C 2x C 3
u2 C 2
2
2
For the second integral, let u D x 2 C 2x C 3. We want du D .2x C 2/ dx to appear in the numerator, so we write
Z
Z
Z
Z
2x dx
.2x C 2 # 2/ dx
.2x C 2/ dx
dx
D
D
#
2
.x 2 C 2x C 3/2
.x 2 C 2x C 3/2
.x 2 C 2x C 3/2
.x 2 C 2x C 3/2
Z
Z
Z
du
dx
1
dx
D
#
2
D
#
#
2
u
u2
.x 2 C 2x C 3/2
.x 2 C 2x C 3/2
Z
#1
dx
D 2
#2
:
x C 2x C 3
.x 2 C 3x C 3/2
Finally, for this last integral, complete the square, then substitute u D x C 1, du D dx:
Z
Z
Z
dx
dx
du
D
D
:
.x 2 C 2x C 3/2
..x C 1/2 C 2/2
.u2 C 2/2
The Method of Partial Fractions
S E C T I O N 7.5
p
p
Now use the trigonometric substitution u D 2 tan !. Then du D 2 sec2 ! d!, and u2 C 2 D 2 tan2 ! C 2 D 2 sec2 !. Thus
p Z
p %
p
p
&
Z
Z p
du
2 sec2 ! d!
2
2 1
1
2
2
2
D
D
cos
!
d!
D
!
C
sin
!
cos
!
D
!
C
sin ! cos ! C C:
2
2
4
4
4 2
2
8
8
.u C 2/
4 sec !
p
We construct a right triangle with tan ! D u= 2:
u2 + 2
u
2
p
p p
From this we see that sin ! D u= u2 C 2 and cos ! D 2= u2 C 2. Therefore
!
p
p
!
" p !
"
Z
du
2
u
2
u
2
!1
D
tan
C
CC
p
p
p
.u2 C 2/2
8
8
2
u2 C 2
u2 C 2
p
p
!
"
!
"
2
u
u
2
xC1
!1
!1 x C 1
D
tan
p
C
C
C
D
tan
p
C
C C:
2 C 2/
2 C 2x C 3/
8
8
4.u
4.x
2
2
Collecting all the terms, we have
Z
!#
p
!
" "
!
"
.x 2 C 3/ dx
1
#1
2
xC1
!1 x C 1
!1 x C 1
D
tan
#
#
2
tan
C
CC
p
p
p
8
.x 2 C 2x C 3/2
x 2 C 2x C 3
4.x 2 C 2x C 3/
2
2
2
p !
!
"
1
2
xC1
2 C .x C 1/
D p C
tan!1
p
C
CC
4
2.x 2 C 2x C 3/
2
2
p
!
"
3 2
xC1
xC3
D
tan!1
p
C
C C:
4
2.x 2 C 2x C 3/
2
In Exercises 45–48, evaluate by using first substitution and then partial fractions if necessary.
Z
x dx
45.
x4 C 1
Use the substitution u D x 2 so that du D 2x dx, and
Z
Z
x
1
1
1
1
dx
D
du D tan!1 u D tan!1 .x 2 /
2
2
2
x4 C 1
u2 C 1
Z
x dx
46.
.x C 2/4
SOLUTION
SOLUTION
47.
Z
Use the substitution u D x C 2 and du D dx; then
Z
Z
Z
Z
x
u#2
1
1
dx
D
du
D
du
#
2
du
.x C 2/4
u4
u3
u4
D#
e x dx
# ex
1
2
2
1
C 3 CC D
#
CC
2u2
3u
3.x C 2/3 2.x C 2/2
e 2x
SOLUTION
Use the substitution u D e x . Then du D e x dx D u dx so that dx D
Z
e x dx
D
e 2x # e x
Z
u ! u1 du
D
u2 # u
Z
1
u
du. Then
1
du
u.u # 1/
Using partial fractions, we have
1
B
.A C B/u # A
A
D
D C
u.u # 1/
u
u#1
u.u # 1/
Upon equating coefficients in the numerators, we have A C B D 0, A D #1 so that B D 1. Then
Z
Z
Z
1
1
e x dx
D
#
C
du
du D ln ju # 1j # ln juj C C D ln je x # 1j # ln e x C C
u
u#1
e 2x # e x
879
880
48.
CHAPTER 7
Z
TECHNIQUES OF INTEGRATION
sec2 ! d!
tan2 ! # 1
SOLUTION
Let u D tan !; then du D sec2 ! d! and
Z
Z p
sec2 ! d!
D
tan2 ! # 1
Z
1
du D #
u2 # 1
Z
1
du D # tanh!1 .u/ C C D # tanh!1 .tan !/ C C
1 # u2
p
x dx
. Hint: Use the substitution u D x (sometimes called a rationalizing substitution).
x #1
p
p
SOLUTION Let u D x. Then du D .1=2 x/ dx D .1=2u/ dx. Thus
49. Evaluate
Z p
Z
Z 2
Z
x dx
u.2u du/
u du
.u2 # 1 C 1/ du
D
D
2
D
2
2
2
x#1
u #1
u #1
u2 # 1
!
Z
Z
Z
Z
u2 # 1
1
2 du
2 du
D2
C
du
D
2
du
C
D
2u
C
:
u2 # 1
u2 # 1
u2 # 1
u2 # 1
The partial fraction decomposition of the remaining integral has the form:
2
2
A
B
D
D
C
:
.u # 1/.u C 1/
u#1
uC1
u2 # 1
Clearing denominators gives us
2 D A.u C 1/ C B.u # 1/:
Setting u D 1 yields 2 D A.2/ C 0 or A D 1, while setting u D #1 yields 2 D 0 C B.#2/ or B D #1. The result is
u2
2
1
#1
D
C
:
u#1
uC1
#1
Thus,
Z
2 du
D
u2 # 1
Z
du
#
u#1
Z
du
D ln ju # 1j # ln ju C 1j C C:
uC1
The final answer is
50. Evaluate
SOLUTION
Z
Z p
p
p
p
x dx
D 2u C ln ju # 1j # ln ju C 1j C C D 2 x C ln j x # 1j # ln j x C 1j C C:
x#1
x 1=2
dx
.
# x 1=3
First use the substitution u D x 1=6 . Then
du D
1 !5=6
x
dx
6
)
6x 5=6 du D dx
)
6u5 du D dx
and we have (using long division)
Z
Z
Z
Z
dx
6u5
u3
1
D
du D 6
du D 6 u2 C u C 1 C
du
u#1
u#1
u3 # u2
x 1=2 # x 1=3
!
"
1 3 1 2
D6
u C u C u C ln ju # 1j C C D 2u3 C 3u2 C 6u C 6 ln ju # 1j C C
3
2
ˇ
ˇ
ˇ
ˇ
D 2x 1=2 C 3x 1=3 C 6x 1=6 C 6 ln ˇx 1=6 # 1ˇ C C
Z
dx
51. Evaluate
in two ways: using partial fractions and using trigonometric substitution. Verify that the two answers agree.
2
x #1
SOLUTION
The partial fraction decomposition has the form:
A
1
1
B
D
D
C
:
.x # 1/.x C 1/
x#1
xC1
x2 # 1
Clearing denominators gives us
1 D A.x C 1/ C B.x # 1/:
The Method of Partial Fractions
S E C T I O N 7.5
881
Setting x D 1, we get 1 D A.2/ or A D 12 ; while setting x D #1, we get 1 D B.#2/ or B D # 12 . The result is
x2
1
# 12
1
D 2 C
:
x#1
xC1
#1
Thus,
Z
dx
1
D
2
x2 # 1
Z
dx
1
#
x#1 2
Z
dx
1
1
D ln jx # 1j # ln jx C 1j C C:
xC1
2
2
Using trigonometric substitution, let x D sec !. Then dx D tan ! sec ! d!, and x 2 # 1 D sec2 ! # 1 D tan2 !. Thus
Z
Z
Z
Z
dx
tan ! sec ! d!
sec ! d!
cos ! d!
D
D
D
tan !
sin ! cos !
x2 # 1
tan2 !
Z
D csc ! d! D ln j csc ! # cot !j C C:
Now we construct a right triangle with sec ! D x:
x
x2 − 1
1
p
p
From this we see that csc ! D x= x 2 # 1 and cot ! D 1= x 2 # 1. Thus
ˇ
ˇ
ˇ
ˇ
Z
ˇ
ˇ
ˇ
ˇ
dx
1
ˇp x
ˇ C C D ln ˇ px # 1 ˇ C C:
D
ln
#
p
ˇ
ˇ
ˇ
ˇ
2
x #1
x2 # 1
x2 # 1
x2 # 1
To check that these two answers agree, we write
ˇr
ˇ
ˇp
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ x #1ˇ
ˇ x # 1 px # 1 ˇ
ˇ x #1 ˇ
1
1
1 ˇˇ x # 1 ˇˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ:
ln jx # 1j # ln jx C 1j D ˇ
D ln ˇ
!p
ˇ D ln ˇ p
ˇ D ln ˇ p
ˇ
ˇ x C 1ˇ
ˇ xC1
2
2
2 x C 1ˇ
x # 1ˇ
x2 # 1
52.
Graph the equation .x # 40/y 2 D 10x.x # 30/ and find the volume of the solid obtained by revolving the region
between the graph and the x-axis for 0 " x " 30 around the x-axis.
SOLUTION
The graph of .x # 40/y 2 D 10x.x # 30/ is shown below
y
20
10
−20
20
30
x
40
Using the disk method, the volume is given by
V D
Z
30
0
2
" r dx D "
Z
30
0
r
10x.x # 30/
x # 40
!2
dx D "
Z
30
0
10x.x # 30/ dx
:
x # 40
To find the anti-derivative, expand the numerator and then use long division:
10x.x # 30/
10x 2 # 300x
4000
D
D 10x C 100 C
:
x # 40
x # 40
x # 40
Thus,
"
Z
30
0
" Z
#
Z 30
Z 30
30
10x.x # 30/ dx
dx
D " 10
x dx C 100
dx C 4000
x # 40
x # 40
0
0
0
#
$ˇ30
ˇ
D " 5x 2 C 100x C 4000 ln jx # 40j ˇ
0
'(
) (
)*
D " 4500 C 3000 C 4000 ln.10/ # 0 C 4000 ln.40/
D .7500 # 4000 ln 4/":
882
TECHNIQUES OF INTEGRATION
CHAPTER 7
In Exercises 53–66, evaluate the integral using the appropriate method or combination of methods covered thus far in the text.
Z
dx
53.
p
x2 4 # x2
SOLUTION
Use the trigonometric substitution x D 2 sin !. Then dx D 2 cos ! d!,
4 # x 2 D 4 # 4 sin2 ! D 4.1 # sin2 !/ D 4 cos2 !;
and
Z
dx
p
D
2
x 4 # x2
Z
2 cos ! d!
D
.4 sin2 !/.2 cos !/
1
4
Z
1
csc2 ! d! D # cot ! C C:
4
Now construct a right triangle with sin ! D x=2:
2
x
4 − x2
From this we see that cot ! D
54.
Z
p
4 # x 2 =x. Thus
Z
dx
1
p
D#
4
x2 4 # x2
p
4 # x2
x
!
CC D #
p
4 # x2
C C:
4x
dx
x.x # 1/2
SOLUTION
Using partial fractions, we first write
1
A
B
C
D C
C
:
x
x #1
x.x # 1/2
.x # 1/2
Clearing denominators gives us
1 D A.x # 1/2 C Bx.x # 1/ C C x:
Setting x D 0 yields
1 D A.1/ C 0 C 0
or
A D 1;
while setting x D 1 yields
1D0C0CC
or
C D 1;
1 D 1 C 2B C 2
or
B D #1:
and setting x D 2 yields
The result is
1
1
#1
1
D C
C
:
x
x#1
x.x # 1/2
.x # 1/2
Thus,
55.
Z
Z
dx
D
x.x # 1/2
Z
dx
#
x
Z
dx
C
x#1
Z
dx
1
D ln jxj # ln jx # 1j #
C C:
x#1
.x # 1/2
cos2 4x dx
SOLUTION
Use the substitution u D 4x, du D 4 dx. Then we have
Z
cos2 .4x/ dx D
D
1
4
Z
cos2 .4x/4 dx D
1
4
Z
cos2 u du D
&
%
1 1
1
u C sin u cos u C C
4 2
2
1
1
1
1
u C sin u cos u C C D x C sin 4x cos 4x C C:
8
8
2
8
The Method of Partial Fractions
S E C T I O N 7.5
56.
Z
x sec2 x dx
SOLUTION
57.
Z
883
Use integration by parts, with u D x and v 0 D sec2 x. Then u0 D 1, v D tan x, and
Z
Z
(
)
x sec2 x dx D x tan x # tan x dx D x tan x # # ln j cos xj C C D x tan x C ln j cos xj C C:
dx
.x 2 C 9/2
SOLUTION
Use the trigonometric substitution x D 3 tan !. Then dx D 3 sec2 ! d!,
x 2 C 9 D 9 tan2 ! C 9 D 9.tan2 ! C 1/ D 9 sec2 !;
and
Z
dx
D
.x 2 C 9/2
Z
3 sec2 ! d!
3
D
81
.9 sec2 !/2
Z
sec2 ! d!
1
D
27
sec4 !
Z
cos2 ! d! D
1
27
!
"
1
1
! C sin ! cos ! C C:
2
2
Now construct a right triangle with tan ! D x=3:
x2 + 9
x
3
p
p
From this we see that sin ! D x= x 2 C 9 and cos ! D 3= x 2 C 9. Thus
!
"!
"
Z
# $
#x$
dx
1
1
x
3
1
x
!1 x
D
tan
C
p
p
CC D
tan!1
C
C C:
p
2
2 C 9/
2C9
2C9
54
3
54
54
3
18.x
2
x
x
x C9
Z
58.
! sec!1 ! d!
SOLUTION
Use Integration by Parts, with u D sec!1 ! and v 0 D !. Then u0 D 1=!
Z
! sec!1 ! d! D
!2
sec!1 ! #
2
Z
p
! 2 # 1, v D ! 2 =2, and
! 2 d!
!2
1
p
D
sec!1 ! #
2
2
2
2! ! # 1
Z
! d!
p
:
!2 # 1
To evaluate the remaining integral, use the substitution w D ! 2 # 1, dw D 2! d!. Then
Z
Z
Z
p
! d!
1
2! d!
1
dw
1( p )
p
D
p
D
p D
2 w C C D ! 2 # 1 C C:
2
2
2
w
!2 # 1
!2 # 1
The final answer is
59.
Z
Z
! sec!1 ! d! D
tan5 x sec x dx
SOLUTION
!2
1p 2
sec!1 ! #
! # 1 C C:
2
2
Use the trigonometric identity tan2 x D sec2 x # 1 to write
Z
Z #
$2
tan5 x sec x dx D
sec2 x # 1 tan x sec x dx:
Now use the substitution u D sec x, du D sec x tan x dx:
Z
Z
Z #
$
tan5 x sec x dx D .u2 # 1/2 du D
u4 # 2u2 C 1 du
60.
Z
D
1 5 2 3
1
2
u # u C u C C D sec5 x # sec3 x C sec x C C:
5
3
5
3
.3x 2 # 1/ dx
x.x 2 # 1/
SOLUTION
The denominator expands to x 3 # x, so if we let u D x 3 # x, then du D .3x 2 # 1/ dx, which is the numerator. Thus
Z
.3x 2 # 1/ dx
D
x.x 2 # 1/
Z
du
D ln juj C C D ln.x.x 2 # 1// C C
u
884
61.
Z
ln.x 4 # 1/ dx
SOLUTION
Apply integration by parts with u D ln.x 4 # 1/, v 0 D 1; then u0 D
Z
62.
TECHNIQUES OF INTEGRATION
CHAPTER 7
Z
.x 2
and v D x, so after simplification,
Z
x4
1
4
dx
D
x
ln.x
#
1/
#
4
1C 4
dx
x4 # 1
x #1
Z
Z
1
D x ln.x 4 # 1/ # 4 1 dx # 4
dx
x4 # 1
!
"
Z
1
1
1
D x ln.x 4 # 1/ # 4x # 4
#
dx
2 x2 # 1 x2 C 1
Z
Z
1
1
D x ln.x 4 # 1/ # 4x # 2
dx
C
2
dx
2
2
x #1
x C1
ln.x 4 # 1/ dx D x ln.x 4 # 1/ # 4
Z
4x 3
x 4 !1
D x ln.x 4 # 1/ # 4x C 2 tanh!1 x C 2 tan!1 x C C
x dx
# 1/3=2
Use the substitution u D x 2 # 1, du D 2x dx. Then we have
Z
Z
Z
x dx
1
2x dx
1
du
1
#1
#1
D
D
D .#2/u!1=2 C C D p C C D p
C C:
2
2
2
u
.x 2 # 1/3=2
.x 2 # 1/3=2
u3=2
x2 # 1
Z
x 2 dx
63.
2
.x # 1/3=2
SOLUTION
SOLUTION
Use the trigonometric substitution x D sec !. Then dx D sec ! tan ! d!,
x 2 # 1 D sec2 ! # 1 D tan2 !;
and
Z
Z
Z
.sec2 !/ sec ! tan ! d!
sec3 ! d!
.tan2 ! C 1/ sec ! d!
D
D
tan2 !
tan2 !
.tan2 !/3=2
Z
Z
Z
Z
tan2 ! sec ! d!
sec ! d!
D
C
D sec ! d! C csc ! cot ! d!
tan2 !
tan2 !
x 2 dx
D
.x 2 # 1/3=2
Z
D ln j sec ! C tan !j # csc ! C C:
Now construct a right triangle with sec ! D x:
x
x2 − 1
1
From this we see that tan ! D
64.
Z
.x C 1/ dx
C 4x C 8/2
.x 2
p
p
x 2 # 1 and csc ! D x= x 2 # 1. So the final answer is
Z
ˇ
ˇ
p
x 2 dx
x
ˇ
ˇ
D ln ˇx C x 2 # 1ˇ # p
C C:
2
.x 2 # 1/3=2
x #1
SOLUTION At first it might appear that one would use partial fractions to simplify this problem, but in fact it’s already in simplified
form. Instead, use the substitution u D x 2 C 4x C 8, du D .2x C 4/ dx. Then we have
Z
Z
Z
.x C 1/ dx
1
.2x C 2/ dx
1
.2x C 2 C 2 # 2/ dx
D
D
2
2
2
2
2
2
.x C 4x C 8/
.x C 4x C 8/
.x 2 C 4x C 8/2
Z
Z
1
.2x C 4/ dx
dx
D
#
2
2
2
2
.x C 4x C 8/
.x C 4x C 8/2
Z
Z
Z
1
du
dx
dx
#1
D
#
#
D
:
2
2
2
2
2
2u
u
.x C 4x C 8/
.x C 4x C 8/2
To evaluate the remaining integral, complete the square, then let w D x C 2, dw D dx:
Z
Z
Z
Z
dx
dx
dx
dw
D
D
D
:
.x 2 C 4x C 8/2
.x 2 C 4x C 4 C 4/2
..x C 2/2 C 4/2
.w 2 C 4/2
The Method of Partial Fractions
S E C T I O N 7.5
885
Next, let w D 2 tan !, dw D 2 sec2 ! d!. Then
w 2 C 4 D 4 tan2 ! C 4 D 4.tan2 ! C 1/ D 4 sec2 !;
and we have
Z
dw
D
.w 2 C 4/2
Z
2 sec2 ! d!
1
1
D cos2 ! d! D
8
8
16 sec4 !
!
"
1
1
1
1
! C sin ! cos ! C C D
!C
sin ! cos ! C C:
2
2
16
16
Now construct a right triangle with tan ! D w=2:
w2 + 4
w
2
p
p
From this we see that sin ! D w= w 2 C 4 and cos ! D 2= w 2 C 4. Thus
!
"!
"
Z
# $
#w $
dw
1
1
w
2
1
w
!1 w
D
tan
C
p
p
CC D
tan!1
C
C C:
2
2
2
2
16
2
16
16
2
.w C 4/
8.w 2 C 4/
w C4
w C4
In terms of x, we have
Z
dx
D
.x 2 C 4x C 8/2
Z
dw
1
D
tan!1
16
.w 2 C 4/2
!
xC2
2
"
C
xC2
C C:
8..x C 2/2 C 4/
Collecting all the terms, we have
!
"
Z
.x C 1/ dx
#1
1
xC2
!1 x C 2
D
#
tan
#
CC
2
.x 2 C 4x C 8/2
2.x 2 C 4x C 8/ 16
8.x 2 C 4x C 8/
!
"
1
xC2
xC6
D # tan!1
#
C C:
16
2
8.x 2 C 4x C 8/
p
Z
x dx
65.
x3 C 1
SOLUTION
66.
Z
Use the substitution u D x 3=2 , du D 32 x 1=2 dx. Then x 3 D .x 3=2 /2 D u2 , so we have
Z p
Z
x dx
2
du
2
2
D
D tan!1 u C C D tan!1 .x 3=2 / C C:
3
3
3
x3 C 1
u2 C 1
x 1=2 dx
x 1=3 C 1
SOLUTION
Use the substitution u D x 1=6 , du D 16 x !5=6 dx. Then dx D 6x 5=6 du D 6u5 du, and we get
Z
x 1=2 dx
D
x 1=3 C 1
Z
u3 .6u5 du/
D6
u2 C 1
Z
u8 du
:
u2 C 1
By long division
u8
1
D u6 # u4 C u2 # 1 C 2
;
u2 C 1
u C1
thus
Z
u8
du D
2
u C1
Z !
u6 # u4 C u2 # 1 C
1
2
u C1
"
du D
1 7 1 5 1 3
u # u C u # u C tan!1 u C C:
7
5
3
The final answer is
Z
x 1=2
6
6
D x 7=6 # x 5=6 C 2x 1=2 # 6x 1=6 C 6 tan!1 .x 1=6 / C C:
1=3
7
5
x
C1
67. Show that the substitution ! D 2 tan!1 t (Figure 1) yields the formulas
cos ! D
1 # t2
;
1 C t2
sin ! D
2t
;
1 C t2
d! D
2 dt
1 C t2
10
This substitution transforms the integral of any rational function Zof cos ! and sin ! into an integral of a rational function of t (which
d!
can then be evaluated using partial fractions). Use it to evaluate
.
cos ! C 34 sin !
886
TECHNIQUES OF INTEGRATION
CHAPTER 7
1 + t2
t
/2
1
FIGURE 1
p
p
If ! D 2 tan!1 t, then d! D 2 dt=.1 C t 2 /. We also have that cos. "2 / D 1= 1 C t 2 and sin. "2 / D t= 1 C t 2 . To
find cos !, we use the double angle identity cos ! D 1 # 2 sin2 . "2 /. This gives us
SOLUTION
!
"2
t
2t 2
1 C t 2 # 2t 2
1 # t2
cos ! D 1 # 2 p
D1#
D
D
:
1 C t2
1 C t2
1 C t2
1 C t2
To find sin !, we use the double angle identity sin ! D 2 sin. "2 / cos. "2 /. This gives us
!
"!
"
t
1
2t
sin ! D 2 p
p
D
:
2
2
1
C
t2
1Ct
1Ct
With these formulas, we have
Z
d!
D
cos ! C .3=4/ sin !
Z
2 dt
#
2
$ 1Ct #
$ D
1!t 2
2t
C 34 1Ct
2
1Ct 2
Z
8 dt
D
4.1 # t 2 / C 3.2t/
Z
8 dt
D
4 C 6t # 4t 2
Z
4 dt
:
2 C 3t # 2t 2
The partial fraction decomposition has the form
4
A
B
D
C
:
2#t
1 C 2t
2 C 3t # 2t 2
Clearing denominators gives us
4 D A.1 C 2t/ C B.2 # t/:
Setting t D 2 then yields
4 D A.5/ C 0
while setting t D # 12 yields
4D0CB
! "
5
2
or
AD
or
4
;
5
BD
8
:
5
The result is
4
8
4
5
5
D
C
:
2#t
1 C 2t
2 C 3t # 2t 2
Thus,
Z
4
4
dt D
5
2 C 3t # 2t 2
Z
dt
8
C
2#t
5
Z
dt
4
4
D # ln j2 # tj C ln j1 C 2tj C C:
1 C 2t
5
5
The original substitution was ! D 2 tan!1 t, which means that t D tan. "2 /. The final answer is then
ˇ
ˇ
! "ˇ
! "ˇ
Z
d!
4 ˇˇ
! ˇˇ 4 ˇˇ
! ˇˇ
D
#
ln
2
#
tan
C
ln
1
C
2
tan
C C:
ˇ
ˇ
ˇ
3
5
2
5
2 ˇ
cos ! C 4 sin !
Z
d!
68. Use the substitution of Exercise 67 to evaluate
.
cos ! C sin !
SOLUTION
Using the substitution ! D 2 tan!1 t, we get
Z
d!
D
cos ! C sin !
Z
2 dt=.1 C t 2 /
D
.1 # t 2 /=.1 C t 2 / C 2t=.1 C t 2 /
Z
2 dt
D #2
1 # t 2 C 2t
The partial fraction decomposition has the form
#2
B
A
p C
p :
D
t 2 # 2t # 1
t #1# 2
t #1C 2
Z
dt
:
t 2 # 2t # 1
S E C T I O N 7.5
The Method of Partial Fractions
887
Clearing denominators gives us
Setting t D 1 C
Z
p
p
2/ C B.t # 1 # 2/:
p
, while setting t D 1 # 2 yields B D p1 . Thus,
#2 D A.t # 1 C
2 then yields A D # p1
2
Z
p
2
Z
p
p
1
dt
1
1
p #p
p D p ln jt # 1 C 2j # p ln jt # 1 # 2j C C
t #1C 2
2
t #1# 2
2
2
# $
ˇ
ˇ
p
ˇ tan " # 1 C 2 ˇ
ˇ
ˇ
1
2
ˇ C C:
# $
D p ln ˇˇ
p
ˇ
2 ˇ tan " # 1 # 2 ˇ
2
d!
1
D p
cos ! C sin !
2
dt
Further Insights and Challenges
69. Prove the general formula
Z
dx
1
x#a
D
ln
CC
.x # a/.x # b/
a#b x #b
where a; b are constants such that a ¤ b.
SOLUTION
The partial fraction decomposition has the form:
1
A
B
D
C
:
.x # a/.x # b/
x#a
x#b
Clearing denominators, we get
1 D A.x # b/ C B.x # a/:
Setting x D a then yields
1 D A.a # b/ C 0
or
AD
1
;
a#b
1 D 0 C B.b # a/
or
BD
1
:
b#a
while setting x D b yields
The result is
1
1
1
D a!b C b!a :
.x # a/.x # b/
x#a
x#b
Thus,
Z
dx
1
D
.x # a/.x # b/
a#b
Z
dx
1
C
x#a
b#a
Z
dx
1
1
D
ln jx # aj C
ln jx # bj C C
x #b
a#b
b#a
ˇx # aˇ
1
1
1
ˇ
ˇ
D
ln jx # aj #
ln jx # bj C C D
ln ˇ
ˇ C C:
a#b
a#b
a#b
x#b
70. The method of partial fractions shows that
Z
ˇ 1 ˇ
ˇ
dx
1 ˇ
D ln ˇx # 1ˇ # ln ˇx C 1ˇ C C
2
2
x2 # 1
The computer algebra system Mathematica evaluates this integral as # tanh!1 x, where tanh!1 x is the inverse hyperbolic tangent
function. Can you reconcile the two answers?
SOLUTION
Let
y D tanh x D
e x # e !x
:
e x C e !x
Solving for x in terms of y, we find
.e x C e !x /y D e x # e !x
e !x .1 C y/ D e x .1 # y/
888
CHAPTER 7
TECHNIQUES OF INTEGRATION
1Cy
1#y
ˇ
ˇ
1 ˇ 1 C y ˇˇ
x D ln ˇˇ
2
1#yˇ
e 2x D
Thus,
tanh!1 x D
so
# tanh!1 x D
as desired.
ˇ
ˇ
1 ˇˇ 1 C x ˇˇ
ln ˇ
;
2
1#xˇ
ˇ
ˇ
1 ˇˇ 1 # x ˇˇ
1
1
ln ˇ
D ln j1 # xj # ln j1 C xj;
2
1Cxˇ
2
2
71. Suppose that Q.x/ D .x # a/.x # b/, where a ¤ b, and let P .x/=Q.x/ be a proper rational function so that
P .x/
A
B
D
C
Q.x/
.x # a/
.x # b/
P .a/
P .b/
and B D 0 .
0
Q .a/
Q .b/
(b) Use this result to find the partial fraction decomposition for P .x/ D 3x # 2 and Q.x/ D x 2 # 4x # 12.
(a) Show that A D
SOLUTION
(a) Clearing denominators gives us
P .x/ D A.x # b/ C B.x # a/:
Setting x D a then yields
P .a/ D A.a # b/ C 0
or
AD
P .a/
;
a#b
P .b/ D 0 C B.b # a/
or
BD
P .b/
:
b#a
while setting x D b yields
Now use the product rule to differentiate Q.x/:
Q0 .x/ D .x # a/.1/ C .1/.x # b/ D x # a C x # b D 2x # a # bI
therefore,
Q0 .a/ D 2a # a # b D a # b
Q0 .b/ D 2b # a # b D b # a
Substituting these into the above results, we find
AD
P .a/
Q0 .a/
and B D
P .b/
:
Q0 .b/
(b) The partial fraction decomposition has the form:
P .x/
3x # 2
3x # 2
A
B
D 2
D
D
C
I
Q.x/
.x # 6/.x C 2/
x#6
xC2
x # 4x # 12
AD
P .6/
3.6/ # 2
16
D
D
D 2I
0
Q .6/
2.6/ # 4
8
BD
3.#2/ # 2
#8
P .#2/
D
D
D 1:
0
Q .#2/
2.#2/ # 4
#8
The result is
3x # 2
1
2
C
:
D
x#6
xC2
x 2 # 4x # 12
S E C T I O N 7.5
The Method of Partial Fractions
889
72. Suppose that Q.x/ D .x # a1 /.x # a2 / ! ! ! .x # an /, where the roots aj are all distinct. Let P .x/=Q.x/ be a proper rational
function so that
P .x/
A1
A2
An
D
C
C !!! C
Q.x/
.x # a1 /
.x # a2 /
.x # an /
(a) Show that Aj D
P .aj /
for j D 1; : : : ; n.
Q0 .aj /
(b) Use this result to find the partial fraction decomposition for P .x/ D 2x 2 # 1, Q.x/ D x 3 # 4x 2 C x C 6 D .x C 1/.x #
2/.x # 3/.
SOLUTION
(a) To differentiate Q.x/, first take the logarithm of both sides, and then differentiate:
(
)
'
*
ln Q.x/ D ln .x # a1 /.x # a2 / ! ! ! .x # an / D ln.x # a1 / C ln.x # a2 / C ! ! ! C ln.x # an /
(
)
d
Q0 .x/
1
1
1
ln Q.x/ D
D
C
C !!! C
dx
Q.x/
x # a1
x # a2
x # an
Multiplying both sides by Q.x/ gives us
%
&
1
1
Q0 .x/ D Q.x/
C !!! C
x # a1
x # an
D .x # a2 /.x # a3 / ! ! ! .x # an / C .x # a1 /.x # a3 / ! ! ! .x # an / C ! ! ! C .x # a1 /.x # a2 / ! ! ! .x # an!1 /:
In other words, the ith product in the formula for Q0 .x/ has the .x # ai / factor removed. This means that
Q0 .aj / D .aj # a1 / ! ! ! .aj # aj !1 /.aj # aj C1 / ! ! ! .aj # an /:
Now clear denominators in the expression for P .x/=Q.x/:
P .x/ D
A1 Q.x/
A2 Q.x/
An Q.x/
C
C !!! C
x # a1
x # a2
x # an
D A1 .x # a2 / ! ! ! .x # an / C .x # a1 /A2 .x # a3 / ! ! ! .x # an / C ! ! ! C .x # a1 /.x # a2 / ! ! ! .x # an!1 /An :
Setting x D aj , we get
P .aj / D .aj # a1 /.aj # a2 / ! ! ! .aj # aj !1 /Aj .aj # aj C1 / ! ! ! .aj # an /;
so that
Aj D
P .aj /
P .aj /
D 0
:
.aj # a1 / ! ! ! .aj # aj !1 /.aj # aj C1 / ! ! ! .aj # an /
Q .aj /
(b) Let P .x/ D 2x 2 # 1 and Q.x/ D .x C 1/.x # 2/.x # 3/, so that Q0 .x/ D 3x 2 # 8x C 1. Then a1 D #1, a2 D 2, and
a3 D 3, so that
A1 D P .#1/=Q0 .#1/ D
1
I
12
7
A2 D P .2/=Q0 .2/ D # I
3
17
A3 D P .3/=Q0 .3/ D
:
4
Thus
P .x/
1
7
17
D
#
C
:
Q.x/
12.x C 1/ 1 D 3.x # 2/
4.x # 3/