Blank B.E., Krantz S.G. Calculus: Single Variable

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According to Step 3, we write the second integral on the right side of (6.3.6) as

2 4x 1 8

dx 5

ðx22Þ

1 4

dx:

We next make the direc t substitution u 5 x 2 2,du5 dx, obtaining

1 4

du 5

arctan





1 C:

We resubstitute for x to obtain

2 4x 1 8

dx 5

arctan



x 2 2



1 C: ð6:3 :8Þ

Substituting the information from lin es (6.3.7) and (6.3.8) into equation (6.3.6), we

have

4x 2 5

2 4x 1 8

dx 5 2lnðx

2 4x 1 8Þ1

arctan



x 2 2



1 C:

Therefore

2 8x

1 20x 2 5

2 4x 1 8

dx 5 x

1 2lnðx

2 4x 1 8Þ1

arctan



x 2 2



1 C: ¥

QUICK QUIZ

1. What indirect substitution is appropriate for

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

dx?

2. What indirect substitution is appropriate for

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 4x

dx?

3. What indirect substitution is appropriate for

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 2

dx?

4. What indirect substitution is appropriate for

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

10 1 2x 1 x

dx?

Answers

1. x 5 tan(θ)2

.x 5

sinðθÞ 3. x 5

ﬃﬃﬃ

secðθÞ 4. x 1 1 5 3 tan(θ)

EXERCISES

Problems for Practice

c In each of Exercises 1210, evaluate the given integral by

making a trigonometric substitution (even if you spot another

way to evaluate the integral). b

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

9 2 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

9 2 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 4x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

25 2 x

ð1 2 x

3=2

ð4 2 x

3=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 4x

10.

ð1 2 9x

3=2

c In each of Exercises 11220, evaluate the given integral by

making

a trigonometric substitution (even if you spot another

way to evaluate the integral). b

6.3 Trigonometric Substitution 495

11.

1 1

12.

12x

1 9

13.

1 1

14.

1 4

15.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 4x

16.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 2x

17.

ð1 1 x

3=2

18.

ð5 1 x

3=2

19.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

20.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

c In each of Exercises 21230, evaluate the given integral by

making

a trigonometric substitution (even if you spot another

way to evaluate the integral). b

21.

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4

22.

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

23.

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

24.

ﬃﬃﬃﬃ

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4

25.

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

26.

ﬃﬃﬃﬃ

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

27.

ﬃﬃﬃﬃ

ﬃﬃ

ðx

2 1Þ

3=2

28.

ﬃﬃ

ðx

2 2Þ

3=2

29.

ﬃﬃ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

30.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

c In each of Exercises 31238, evaluate the given integral. b

31.

2x 1 3

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

32.

1 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

33.

2x 1 3

1 1

34.

1 2

1 1

35.

1 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1

36.

1 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 4

37.

1 1

ð92x

3=2

38.

1 4x 1 3

ð124x

3=2

Further Theory and Practice

c Each of the integrands in Exercises 39246 involves an

expression of the form a

2 b

, a

1 b

,orb

2 a

. Use

an indirect substitution of the form x 5 (a/b) sin(θ), x 5 (a/b)

tan(θ), or x 5 (a/b) sec (θ) to calculate the given integral. b

39.

ﬃﬃ

16x

1 8

40.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

25 2 16x

41.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

16 1 9x

42.

ﬃﬃ

4=3

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 16

43.

ð9 1 4x

3=2

44.

4=5

ð25x

2 16Þ

1=2

45.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

25x

2 16

46.

16x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

9 2 4x

c In each of Exercises 47250, calculate the given integral by

ﬁrst

integrating by parts and then making a trigonometric

substitution. b

47.

ﬃﬃ

arcsecðxÞdx

48.

ﬃﬃ

xarcsin ðxÞdx

49.

1=2

arcsinðxÞdx

50.

ﬃﬃ

arcsecðxÞdx

c In Exercises 51258, calculate the given integral. b

51.

ðx

1 1Þ

52.

2 1

ðx

1 1Þ

53.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

54.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 1

496 Chapter 6 Techniques of Integration

55.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

56.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1

57.

ð1 1 x

3=2

58.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

c In Exercises 59274, calculate the given integral. b

59.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2x 2 x

60.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2x 1 2

61.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2x 1 2

62.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 x 2 1

63.

ðx

22x12Þ

64.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 2x 2 3

65.

2 14x 1 58

66.

1 2x 1 5

67.

2 6x 1 8

1 4

68.

2x 1 4

1 4x 1 13

69.

12x

1 6x 1 5

70.

1 20x 2 3

1 4x 1 13

71.

ðx

1 2x 1 2Þ

72.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 4x 2 5

73.

x 1 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

10x 2 x

74.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

16x 2 x

2 48

c In each of Exercises 75278, divide before integrating. b

75.

1 1 x

4 1 x

76.

1 x

9 1 x

77.

1 1 x

5 1 4x 1 x

78.

1 1

5 2 2x 1 x

c In each of Exercises 79284, make a (nontrigonometric)

direct substitution x 5 φ(u) to evaluate the given integral. b

79.

1 1

ﬃﬃﬃ

80.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1

ﬃﬃﬃ

81.

1 2

ﬃﬃﬃ

1 1

ﬃﬃﬃ

82.

1=4

1 1

1=2

1 1

83.

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

84.

1 1 x

1=3

85. Suppose that a is a positive constant. Show that the sub-

stitution x 5 a sin(θ) leads to the formula

2 x

dx 5



x 1 a

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 x



1 C:

Show that the substitution x 5 a sec(θ) leads to the

formula

2 x

dx 5



x 1 a

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 a



1 C:

Use these formulas to evaluate α 5

1=2

12x

dx and

β 5

ﬃﬃ

1 2 x

dx. Show that the identity

2 x



x 1 a

x 2 a



leads to the formula

2 x

dx 5





x 1 a

x 2 a





1 C

and use it to verify your answers for α and β.

Calculator/Computer Exercises

86. Let b be the abscissa of the point of intersection of the

curves y 5 2x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

and y 5 x in the ﬁrst quadrant.

Find the area of the region between the two curves for

0 # x # b.

87. Find the area of the region that is bounded above by

y 5 ð1 1 x

23=2

and below by y 5 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

88. Let a be the smaller abscissa, b the larger, of the two ﬁrst-

quadrant points of intersection of the curves

y 5 3 2 x=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

and y 5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2 x

=x. Find the area of the

region between the two curves for a # x # b.

89. Let b be the abscissa of the point of intersection of the

curves y 5 2x 2 x

and y 5 x

=ð2 1 2x 1 x

Þ in the ﬁrst

quadrant. Find the area of the region between the two

curves for 0 # x # b.

90. Let a be the negative x-intercept of y 5

2 2x 1 1

2 2x 1 2

. Find

the total area of the regions between the curve and the

x-axis for a # x # 1.

6.3 Trigonometric Substitution 497

6.4 Partial Fractions—Linear Factors

In this section, you will begin to learn how to integrate quotients of polynomials;

such functions are called rational functions. The basic scheme, called partial frac-

tions, consists of writing any rational function as a sum of building blocks, which we

already know how to integrate. Let us now describe what some of these building

blocks are.

Simple Linear Building Blocks

x 2 a

;

where A and a are constants. The integral of this building block is

A  lnðjx 2 ajÞ:

Repeated Linear Building Blocks

ðx 2 aÞ

;

where m is an integer greater than 1. The integral of such a building block is

m 2 1



ðx 2 aÞ

m21

These are all the linear building blocks, and you can see that they are easy to

understand. A great many rational functions can be written as a sum of linear

building blocks and repeated linear building blocks. Let us see how this is done.

The Method of Partial

Fractions for Linear

Factors

⁄ EXAMPLE 1 Integrate

ðx 2 1Þðx 1 2Þ

dx:

Solution The

scheme is to rew rite the integrand as

ðx 2 1Þðx 1 2Þ

x 2 1

x 1 2

Here A and B are constants that we must determine. We put the terms on the right

over a common denominator:

ðx 2 1Þðx 1 2Þ

Aðx 1 2Þ1 Bðx 2 1Þ

ðx 2 1Þðx 1 2Þ

Both sides of this last equation have the same denominator. The only way that

equality can hold is if the numerators are equal:

3 5 Aðx 1 2Þ1 Bðx 2 1Þ: ð6:4:1Þ

498 Chapter 6 Techniques of Integration

Equation (6.4.1) can be reorganized to more clearly reﬂect its status as an identity

of polynomials in x:

0x 1 3 5 ðA 1 BÞx 1 ð2A 2 BÞ:

For two polynomials to be equal, their constant terms must be equal, their linear

terms must be equal, and so on. Thus

3 5 2A 2 B and 0 5 A 1 B:

We solve these two linear equations simultaneously. The second equation tells us

that B 52A; substituting this value for B in the ﬁrst equation results in 3 5 2A 2

(2A), or 3 5 3A. Thus A 5 1 and B 521. We have discovered that

ðx 2 1Þðx 1 2Þ

x 2 1

2 1

x 1 2

Therefore

ðx 2 1Þðx 1 2Þ

dx 5

x 2 1

dx 2

x 1 2

5 lnðjx 2 1jÞ2 lnðjx 1 2jÞ1 C

5 ln





x 2 1

x 1 2





1 C: ¥

INSIGHT

Itisimportanttounderstandthatwedonotsolveequation(6.4.1)forx.

The procedure is to ﬁnd values for A and B that convert equation (6.4.1) into an identity in x.

In other words, we are to ﬁnd A and B so that equation (6.4.1) is valid for every x.

Substituting any value for x in equation (6.4.1) results in an equation in the two unknowns A

and B. This leads to a very simple alternative method: We solve for each unknown by

choosing a value of x that eliminates the other unknown. Thus when x 5 1, the coefﬁcient of

B in (6.4.1) is 0, and the equation reduces to 3 5 3A or A 5 1. Letting x be 22resultsin

3 523B or B 521. Notice that there is no need to solve simultaneous equations with

this technique.

The process used in Example 1 relies on a general algebraic technique for

handling rational functions, which we now begin to describe. In the ﬁrst case that

we discuss, we assume that the denominator factors as (x 2 a

)(x 2 a

) ... (x 2 a

)

with no repetition among the roots a

, a

,...,a

The Method of Partial

Fractions for Distinct

Linear Factors

To integrate a function of the form

pðxÞ

ðx 2 a

Þðx 2 a

Þ  ðx 2 a

;

where p(x) is a polynomial and the a

are distinct real numbers, ﬁrst perform the

following two algebraic steps:

1. Be sure that the degree of p is less than the degree of the denominator; if it is

not, divide the denominator into the numerator. This step must be performed

even if the degrees of the denominator and the numerator are equal.

2. Decompose the function into the form

x 2 a

1 1

x 2 a

;

6.4 Partial Fractions—Linear Factors 499

solving for the numerators A

, A

,...,A

. The result is called the partial fraction

decomposition of the original rational function.

⁄ EX

AMPLE 2 Calculate the integral

2 9x

2 5x 1 9

2 4x 2 5

dx:

Solution We

ﬁrst notice that the degree of the numerator is not less than the

degree of the denominator, so we must divi de. We ﬁnd that

2 9x

2 5x 1 9

2 4x 2 5

5 2x 2 1 1

x 1 4

2 4x 2 5

5 2x 2 1 1

x 1 4

ðx 1 1Þðx 2 5Þ

: ð6:4:2Þ

Observe that the rational function that ends line (6.4.2) has a numerator with

degree less than the degree of the denominator an d a denominator that is the

product of distinct linear factors. We may therefore apply the method of partial

fractions for distinct linear factors. The next step is to solve for the unknowns A

and B in the decomposition

x 1 4

ðx 1 1Þðx 2 5Þ

x 1 1

x 2 5

Putting the right side over a common denominator

x 1 4

ðx 1 1Þðx 2 5Þ

Aðx 2 5Þ1 Bðx 1 1Þ

ðx 1 1Þðx 2 5Þ

and equating numera tors leads to

x 1 4 5 Aðx 2 5Þ1 Bðx 1 1Þ: ð6:4:3Þ

Let us continue with the simple method that we described in the preceding Insight.

One at a time, we substitute x 521 and x 5 5 into equation (6.4.3). The substitu-

tion x 521 results in 21 1 4 5 A(21 2 5) 1B  0, or 3 526A,orA 521/2. The

substitution x 5 5 results in 5 1 4 5 A  0 1 B(5 1 1), or 9 5 6B,o

rB 5 3/2. In

conclusion,

2 9x

2 5x 1 9

2 4x 2 5

dx 5



2x 2 1 1

ð21=2Þ

x 1 1

3=2

x 2 5



ð2x 2 1Þdx 1

21=2

x 1 1

dx 1

3=2

x 2 5

5 x

2 x 2

lnðjx 1 1jÞ1

lnðjx 2 5jÞ1 C: ¥

INSIGHT

Notice the way the values of x are chosen in Example 2. We choose

x 521 so that the coefﬁcient of B will be 0. The result is an equation that involves only

A. Similarly, the choice x 5 5 eliminates the unknown A, leaving us with an equation in B

alone. As in the Insight to Example 1, the values substituted for x are the roots of the

original denominator.

500 Chapter 6 Techniques of Integration

Heaviside’s Method In Examples 1 and 2, we described a shortcut for determining the partial fraction

decomposition

pðxÞ

ðx 2 a

Þðx 2 a

Þ  ðx 2 a

x 2 a

1 1

x 2 a

ð6:4:4Þ

when a

, a

, ... ,a

are distinct real numbers. In that method, we evaluate a suitable

expression at each of the roots a

, a

,...,a

. The idea is that we can pick off the

coefﬁcients A

, A

,...,A

without solving multiple equations simultaneously. Of

course, we cannot set x 5 a

in equation (6.4.4) itself because that would involve

division by 0. It turns out that there is an efﬁcient algorithm for determining equa-

tions that are suitable for the substitutions x 5 a

. The procedure is named after the

electrical engineer Oliver Heaviside (18501925), who used it systematically in his

work. Let q

(x) denote what is left of the denominator (x 2 a

)(x 2 a

) (x 2 a

)

after the factor (x 2 a

) is removed. To determine the coefﬁcient A

, simply multiply

both sides of equation (6.4.4) by x 2 a

. This results in

pðxÞ

ðxÞ

5 A

1 ðx 2 a

Þðterms that involve the other coefficientsÞ:

Now set x 5 a

.Weseeimmediatelythat

pða

ða

: ð6:4:5Þ

⁄ EX

AMPLE 3 Use Heaviside’s method to ﬁnd the partial fraction decom-

position for

1 x 2 1

xðx 2 3Þðx 1 2Þ

Solution We

note that the numerator has smaller degree than the denominator

and that the denominator factors into three distinc t linear terms. We may proceed

with Heaviside’s metho d. We write

1 x 2 1

xðx 2 3 Þðx 1 2Þ

x 2 0

x 2 3

x 2 ð22Þ

: ð6:4:6Þ

The roots of the denominator polynomial are a

5 0, a

5 3, and a

522. We let

p(x) denote the numerator 3x

1 x 2 1. To determine A, multiply both sides by x.

The result is

1 x 2 1

ðx 2 3Þðx 1 2Þ

5 A 1 x 



x 2 3

x 1 2



Now we substitute x 5 0 into this equation to ﬁnd that A 521/(26). To ﬁnd B,

multiply equation (6.4.6) through by x 2 3. The result is

1 x 2 1

xðx 1 2Þ

5 B 1 ðx 2 3Þ



x 1 2



Now substituting in x 5 3 yields B 5 29/15. A similar computation shows that C 5 9/10.

Try it!

INSIGHT

After a little practice, you will be able to carry out Heaviside’s method

with ease. If you had looked at the ﬁnal decomposition of Example 5 without seeing the

calculation, then you might have thought tedious algebra was required to obtain

6.4 Partial Fractions—Linear Factors 501

coefﬁcients such as 1/6, 29/15, and 9/10. But, as we have seen, Heaviside’s method is as

easy to apply when the coefﬁcients are “messy” as it is when the coefﬁcients are simple

integers. Remember, however, that the procedure we have described works only when

the denominator of the integrand factors into distinct linear factors.

Repeated Linear

Factors

When one or more of the linear factors in the denominator is repeated, then we

must use a slightly different partial fracti ons scheme.

Partial Fractions When Repeated Linear Factors Are Present

Consider the rational function

pðxÞ

ðx 2 a

ðx 2 a

;

where p(x) is a polynomial, the a

are distinct real numbers, and the m

are positive integers (some of them

exceeding 1). To integrate a function of this type, perform the following two steps:

1. Be sure that the degree of p is less than the degree of the denominator; if it is not, divide the denominator into

the numerator.

2. For each of the factors (x 2 a

)

in the denominator of the rational function being considered, the partial

fraction decomposition must contain terms of the form

ðx 2 a

1 1

ðx 2 a

⁄ EX

AMPLE 4 Evaluate the integral

1 18x 2 1

ðx 1 4Þ

ðx 2 3Þ

dx:

Solution We

must decompose the integrand into the form

1 18x 2 1

ðx 1 4Þ

ðx 2 3Þ

x 1 4

ðx 1 4Þ

x 2 3

Notice that, because (x 1 4) is a repeated factor in the denominator of the

integrand, our partial fraction decomposition must contain all powers of (x 1 4)

up to and including the highest power occurring in the denominator of the

integrand.

Now, putting the right-hand side over a common denominator (and equating

numerators), leads to

1 18x 2 1 5 A

ðx 1 4Þðx 2 3Þ1 A

ðx 2 3Þ1 Bðx 1 4Þ

Equating the coefﬁcients of like powers of x gives the system

1 B 5 5

1 A

1 8B 5 18

2 12A

2 3A

1 16B 521 :

502 Chapter 6 Techniques of Integration

The ﬁrst equation allows us to eliminate B 5 5 2 A

from the remaining two

equations. They become

27A

1 A

5222 and 228A

23A

5281:

We solve these equations in standard fashio n to obtain A

5 3andA

521. Finally

B 5 5 2 A

5 2. Thus

1 18x 2 1

ðx 1 4Þ

ðx 2 3Þ

dx 5

x 1 4

dx 1

ð21Þ

ðx 1 4Þ

dx 1

x 2 3

dx 5 3lnðjx 1 4jÞ1

ðx 1 4Þ

1 2lnðjx 2 3jÞ1 C:

INSIGHT

In Example 4, you might be tempted to leave out one of the terms A

/(x 1 4)

or A

/(x 1 4)

on the right-hand side. Try it (Exercises 65 and 66). You will ﬁnd that you are

unable to solve the resulting algebraic system for the coefﬁcients. The more complicated

partial fraction decomposition is really necessary. Algebraic theory tells us that the coefﬁ-

cients of a partial fractions expansion are unique. You cannot succeed when you start out

with the wrong form—unless an omitted coefﬁcient just happens to be 0.

Summary of Basic

Partial Fraction Forms

We conclude by summarizing the techniques of this section with several quick

instances of the different partial fraction form s that can arise:

3x 2 1

ðx 2 4Þðx 1 7Þ

x 2 4

x 1 7

1 x 1 1

xðx

2 1Þ

x 1 1

x 2 1

x 1 3

ðx 2 2Þ

ðx 1 1Þ

x 2 2

ðx 2 2Þ

x 1 1

2 2x 2 1

ðx 1 8Þ

ðx 2 1Þ

x 1 8

ðx 1 8Þ

x 2 1

ðx 2 1Þ

When you begin to do the exercises, refer to these examples to tell which form of

the partial fractions technique you should be using. Also notice this unifying theme:

The number of unknown coefﬁcients in any partial fractions problem is equal to the

degree of the denominator.

QUICK QUIZ

1. True or false: If q(x) is a polynomial of degree n that facto rs into linear terms

and if p(x) is a polynomial of degree m with m , n, then an explicitl y calculated

partial fraction decomposition of p(x)/q(x) requires the determination of n

unknown constants.

2. For what values of A, B, and C is A(x 1 1)(x 1 2) 1 Bx(x 1 2) 1 Cx(x 1 1) 5

1 11x 1 4 an identity in x?

3. What is the form of the partial fraction decomposition of

1 3

xðx

2 4Þ

4. What is the form of the partial fraction decomposition of

1 1

ðx 1 1Þ

Answers

1. True 2. A 5 2, B 5 3, C 521

x 2 2

x 1 2

x 1 1

6.4 Partial Fractions—Linear Factors 503

EXERCISES

Problems for Practice

c In each of Exercises 1210, write down the form of the

partial fraction decomposition of the given rational function.

Do not explicitly calculate the coefﬁcients. b

x 1 2

xðx 1 1Þ

2x 1 3

2 1

1 1

2 4

ðx 2 2Þðx 1 5Þ

ð2x 2 5Þðx 1 4Þ

1 2x 1 2

ðx 2 4Þ

ðx 1 2Þ

1 7

ðx 1 1Þðx 2 2Þ

1 1

ðx 1 1Þ

2 17

ðx 2 5Þ

ðx 1 3Þ

10.

ðx 2 7Þ

ðx 1 1Þ

c In each of Exercises 11220, decompose the given rational

function

into partial fractions. Calculate the coefﬁcients. b

11.

xðx 1 1Þ

12.

5 2 x

2 1

13.

x 1 6

2 4

14.

1 3x 2 10

15.

11x

ð2x 2 5Þðx 1 3Þ

16.

ðx 2 4Þ

ðx 1 2Þ

17.

1 x

2 5x 1 2

ðx 1 1Þðx 2 2Þ

18.

ðx

1 xÞ

19.

2 4x

2 13x 1 76

ðx

2 4x 1 4Þðx

1 6x 1 9Þ

20.

ðx 2 1Þ

ðx 1 1Þ

c In each of Exercises 21226, use Heaviside’s method to

calculate

the partial fraction decomposition of the given

rational function. b

21.

x 1 1

ð2x 1 7Þð2x 1 9Þ

22.

2x 1 1

ðx 2 2Þðx 1 1Þ

23.

1 3x 1 1

ðx 2 2Þðx 1 3Þðx 1 4Þ

24.

1 8x 1 13

ðx 1 1Þðx 1 3Þðx 1 5Þ

25.

1 9x 1 5

2 x

26.

1 5x

1 7x 1 7

ðx 2 1Þxðx 1 1Þðx 1 2Þ

c In each of Exercises 27238, use the method of partial

fractions

to calculate the given integral. b

27.

3x 1 1

2 1

28.

3x 1 4

1 x 2 6

29.

9x 1 18

ðx 2 3Þðx 1 6Þ

30.

2 2x

2 2x 2 2

2 x

31.

1 2x 1 1

32.

1 7x 2 2

xðx 2 1Þðx 1 2Þ

33.

1 14x 1 39

ðx 1 2Þðx 1 3Þð x 1 5Þ

34.

1 4x 1 5

2 x

35.

1 1

1 x

36.

ðx 2 2Þ

ðx 1 1Þ

37.

5x 2 6

ðx 2 2Þ

ðx 1 2Þ

38.

ðx 1 1Þðx24Þ

c Calculate each of the deﬁnite integrals in Exercises

39244. b

39.

4x 2 7

ðx 2 1Þðx 2 4Þ

40.

2x 1 1

xðx 1 1Þ

41.

x 1 6

ðx 2 2Þðx 1 2Þ

504 Chapter 6 Techniques of Integration