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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