Chapter 5

New Words

Rational function 有理函数

Partial fraction 部分分式 Polynomial 多项式

Factorization 因式分解式

Proper fraction 真分式 Factor 分解因式

Improper fraction 假分式

Reducible 可约的 Denominator 分母

Irreducible 不可约的 Numerator 分子

1 How to integrate certain rational functions

spolynomial

are and when .,)(

)( function,

rationalany integrate tohow showssection This

xQxPxQ

xP

At first, we will rewrite any rational function in a

much simpler form

(1) The form of rational function

).integer enonnegativ are ,;,;0,0(

)(

)(

00

110

110

nmRbaba

bxbxb

axaxa

xQ

xP

ii

mmm

nnn

fractionproper a isit , when

fractionimproper an called isit , when

mn

mn

(2) The properties of rational function

a. Any rational improper fraction can be expressed as

the sum of a polynomial and a rational proper

fraction.

b. Any polynomial function Q(x) with real coefficients

can factorize into the product of first-degree polynomial

and the second-degree irreducible polynomials in R

frations-partial simple of kindsfour

following theof sum theas written becan )(

)(

fractionproper rationalany 2, and 1 stepAfter c.

xQ

xP

;)(

II. ;I.kax

A

ax

A

.)(

IV. ;III. 22 kqpxx

NMx

qpxx

NMx

xQxP

function rationalproper a ngRepresenti 3

The general procedure for representing a proper

rational function consists of four steps:

spolynomial

degree-second eirreducibl and spolynomial

degree-first ofproduct a as Writea). xQ

.)()(

sum theformthen ),(

ofion factorizat in the appears )( If c).

22222

211

2

kkk

k

qpxx

NxM

qpxx

NxM

qpxx

NxM

xQ

qpxx

later. determined be toare constants thewhere

;)()(

form

, )( ofion factorizat in the appears )( If b).

221

k

kk

k

A

ax

A

ax

A

ax

A

xQax

later. determined

be toare and constants thewhere kk NM

xQxP

equals 3 and 2 steps

in formed sexpression theall of sum that theso

3 and 2 stepsin mentioned constants all Find d).

(4) Integrate a rational function

Next, we focus on integrating the four types of

function I, II, III, IV. Once we know how to integrate

them, we can integrate any rational function

,ln ,first type The Caxax

dx

),1( ))(1(

1

)(

typesecond The

1

kCaxkax

dxkk

dxqpxx

M

Nx

Mdxqpxx

NMx

22

22

2

type, thirdThe

qpxx

dxMpN

qpxx

qpxxdM

dxqpxx

pM

Npx

M

22

2

2

)2

()(

2

)2

()2(

2

,

4

2arctan

4

1)

2(ln

2

)4

()2

(

)2

(ln2

22

2

22

2

Cp

q

px

pq

MpNqpxx

M

pq

px

dxMpNqpxx

M

kk

k

qpxx

dxMpN

qpxx

qpxxdM

dxqpxx

NMx

)()

2(

)(

)(

2

)(

,fouth type The

22

2

2

.

)]4

()2

[(

)2

(

))(1(

1

2

22

12

k

k

pq

px

dxMpN

qpxxk

M

Example 1

3872

13323928112

oftion representafraction -partial theFind

23

2345

xxx

xxxxx

xQ

xP

Solution

Since the degree of the numerator is greater than

the degree of the denominator, it is improper.

Divide:

42

924216

1326206

616144

3236204

3872

32133239281123872

2

23

23

234

234

2345

2

234523

xx

xxx

xxx

xxxx

xxxx

xxxx

xxxxxxxxxx

3872

4232 , Hence

23

22

xxx

xxxx

xQ

xP

3872

42

oftion representafraction -partial thefind then We

23

2

xxx

xx

3213872

r denominato thefactoringFirst 223 xxxxx

2321

2

32

212

2

13232142

givesr denominato theClearing

3211321

42

Then

xCxCxxCxx

x

C

x

C

x

C

xx

xx

5,1,2 gives

332252

42

sidesboth on tscoefficien Comparing

321

3213212

31

2

CCC

CCCxCCCxCC

xx

32

5

1

1

1

2

321

42

Thus

22

2

xxxxx

xx

Remark:

The above method is called comparison of coefficientscomparison of coefficients.

It depends on the fact that if two polynomials are equal,

then their corresponding coefficients are equal.

The next example illustrates a way for finding the

partial fractions when the factorization of the

denominator involves only linear factors and none of

them is repeated.

Example 2 .d2

32 Find

23x

xxx

x

Solution

21)2)(1(

32

Then

x

C

x

B

x

A

xxx

x

)2)(1(2

:rdenominato factor theFirst 23 xxxxxx

,)2)(1(

)1()2()2)(1(

xxx

xCxxBxxxA

).1()2()2)(1(32

rdenominato Clear the

xCxxBxxxAx

,)2(6

1)1(3

523

232

23

xxxxxxx

6

1,2 ,

3

5,1 ,

2

3 ,0

for holdsit particularIn . allfor holds above The

CxBxAx

x

dx

xxxdx

xxxx

])2(6

1)1(3

523

[2

3223

Cxxx 2ln61

1ln35

ln23

Example 3

Solution

.4

4 Find

3dxxx

,)4(

44

423

xxxx

)4(

)()4(

4)4(

4

Then

2

2

22

xx

CBxxxA

x

CBx

x

A

xx

If a linear factor is repeated you may use either

substitution or comparison coefficients.

,)4(

4)(2

2

xxACxxBA

0

1

1

44

0

0

:equations

following theabovegives theofnumerator

theof sidesboth on the tscoefficien Comparing

C

B

A

A

C

BA

,4

14

423

x

xxxx

dxx

xx

dxxx

)4

1(

44

23

4

)4(211

2

2

xxd

dxx

.4ln21

ln 2 Cxx

Example 4

Solution

.)1(

1 Find

3

3

dxxx

x

,)1(

)1()1()1(

)1()1(1)1(

1

3

23

323

3

xx

DxxCxxBxxA

x

D

x

C

x

B

x

A

xx

x

Clearing the denominator gives

,)1()1()1(1 233 DxxCxxBxxAx

onSubstituti ,2,1 ,1 ,0

for holdsit particularIn . allfor holds above The

DxAx

x

1

2

023

1

tscoefficien comparing Next,

C

B

CBA

BA

,)1(

2)1(

11

21)1(

1323

3

xxxxxxx

Remark

(2) But we can also use other methods to integrate

rational functions.

dxxxxx

dxxx

x

]

)1(2

)1(1

121

[)1(

1323

3

.)1(

21

11ln2ln 2 C

xxxx

(1) Either the substitution method or the comparison o

f coefficients method are used in the above example.

Example 5

Solution

.1

13 Find

3

2

dxxx

x

1)1(

113

3

3

3

2

xxxxd

dxxx

x.1ln 3 Cxx

Example 6

Solution

.d)1(

Find10

3

xx

x

dtt

tdx

xx tx

10

31

10

3 )1()1(

dtt

ttt

10

23 133

dttttt )33( 10987

Ctttt 9876

91

83

73

61

.)1(9

1)1(8

3)1(7

3)1(6

19876

1

Cxxxx

tx

2. How to integrate rational trigonometric functions

So far in this chapter you have met three techniques for

computing integrals. The first, integration of substitution

and the second, integration by parts, are used most often.

Partial fractions applies only to a special class of

integrands, the rational functions. In this section we

compute some rational trigonometric functions.

xnxmxxxxR dcossind)cos,(sin Computing (1)

:identitis theseof aid with the

computed becan integrals threeThese series.Fourier

ofstudy in the needed are dcoscos

,dsincos ,dsinsin integrals The

xnxmx

xnxmxxnxmx

yxyxyx

yxyxyx

yxyxyx

cos2

1cos

2

1coscos

sin2

1sin

2

1cossin

cos2

1cos

2

1sinsin

Example 7

Solution

.2cos3cos Find xdxx

dxxxxdxx )5cos(cos21

2cos3cos

.5sin101

sin21

Cxx

xxxxx

xxxxxR

nmm

m

dcossinor dcosor

dsind)cos,(sin Computing (2)

formulas rictrigonomet

multipleor 1cossin identities the

of aid with thecomputed becan integrals These22 xx

Example 8

Solution

.cossin Find 67 dxxx

xdxxdxxx coscossincossin 6667

xdxx coscos)cos1( 632

.cos10

1cos

9

3cos

8

3cos

7

1

cosd)coscos3cos3cos(

cosdcos)coscos3cos31(

10987

9876

632

Cxxxx

xxxxx

xxxxx

Example 9

Solution

.cossin Find 24 dxxx

xxxx

xxxxxx

dsindsin

d)sin1(sindcossin

64

2424

xxxx

xxx

xx

xx

d)2cos2cos32cos31(8

1

d)2cos2cos21(4

1

d)2

2cos1(d)

2

2cos1(

32

2

32

xxxR d)cos,(sinfor on substituti angle-halfA (3)

by the offunction rational a into dtransforme

is sin and cos offunction rationalAny

u

xx

.2

tan The .2

tanon substituti

xu

xu

fractions. partialby

done becan that integralan yieldson substituti

.arctan2 then ,2

tanLet

,

2tan1

2tan1

cos,

2tan1

2tan2

sin2

2

2

uxx

u

x

x

xx

x

x

2

2

2

2

1

21

1cos

1

2sin

u

dudu

u

ux

u

ux

A half-angle substitution

.d1

2)

1

1,

1

2(d)cos,(sin

Thus,

22

2

2

2tan

uuu

u

u

uRxxxR

xu

Example 10

Solution

.dcos3

sin3 Find x

x

x

xxx

x

xx

x

xx

xx

xx

x

x

cos3lndcos3

3cos3

)cos3(dd

cos3

3

dcos3

sind

cos3

3d

cos3

sin3

Cx

Cu

uu

uu

u

ux

x

xu

)2

tan2

1arctan(

2

32

arctan2

3d

2

3

d1

2

1

13

3d

cos3

3 and

2

2

2

2

2tan

Cxx

dxxx

cos3ln)

2tan

21

arctan(2

3cos3sin3

3. How to integrate simple irrational functions

Some simple irrational functions can be transformed

into a rational function by proper substitution.

For example:

xaxf

xaxf

xxaf

d)(

d)(

d)(

)1(

22

22

22

taxtax

taxtax

taxtax

cscor sec

cotor tan

cosor sin

Let

xxxf

xbaxf

nm

n

d),(

d)()2(

nmptx

tbaxp

n

, of multiplecommon of minimum is ,Let

dxcbxaxx

dxcbxaxx

22

2

1

1

)3(t

x1

Let

dxbaxxR n ),()4( tbaxn Let

dxdcxbax

xR n

),()5( tdcxbax

n Let

Example 11

Solution

.1

Find xx

dx

.d2d,1then ,1Let 2 ttxtxtx

ttt

t

xx

xd

1

2

1

d2

222

2

22

)2

3()

2

1(

)2

1(

1

)1(

d1

1d

1

12

t

td

tt

ttd

ttt

ttt

t

Ct

tt 3

12arctan

32

1ln 2

Cx

xx 3

112arctan

32

1ln

Example 12

Solution

.)2()1(

Find3 2

xx

dx

,21

11

)2()1(3

3 2dx

xx

xxx

dx

,)1(

9,

1

21then ,

2

1Let

23

2

3

3

3 dtt

tdx

t

txt

x

x

dttxx

dx

33 2 13

)2()1(dt

ttt

t

)1

21

1( 2

Ct

ttt 3

12arctan31ln

21

1ln 2

Cxx

xx

xx

xx

3

121

2arctan3

121

)21

(ln21

21

1ln

3

33 23

22

2

)23

()21

(23

112

21

1t

dtdt

ttt

tdt