a rational function is by definition the quotient of two polynomial functions, which has the...
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![Page 1: A rational function is by definition the quotient of two polynomial functions, which has the following form: §4 有理函数的积分 Integration of Rational Function](https://reader038.vdocuments.site/reader038/viewer/2022112000/56649f385503460f94c54cf3/html5/thumbnails/1.jpg)
A rational function is by definition the quotient of
two polynomial functions, which has the following form:
mmmm
nnnn
bxbxbxbaxaxaxa
xQxP
11
10
11
10
)()(
where m and n are nonnegative integers,
naaa ,,, 10 and mbbb ,,, 10 are real numbers,
and 00 a , 00 b .
§4 有理函数的积分 Integration of Rational Function
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(1) ,While n m
the rational function is an proper fraction ;(2) ,While n m
the rational function is an improper fraction.
If there is no common factor between numerator and denominator
Using the polynomial division, the improper fraction can change to the sum of a polynomial and proper fraction.
Eg.1
12
5
x
xx.
1
122
3
x
xxx
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Difficulty:
Chang the proper fraction to the sum of fractions.
General rules for changing the proper fraction to the sum of fractions :
( 1 ) If the denominator has a factor ,
then disassemble
kax )(
,)()( 1
21
axA
axA
axA k
kk
Where kAAA ,,, 21 are constants.
Especially : ,1k disassemble as ;ax
A
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( 2 ) If the denominator has a factor ,kqpxx )( 2
, then disassemble 2 4 0where p q
qpxxNxM
qpxxNxM
qpxxNxM kk
kk
21222
211
)()(
Where ii NM , are constants ),,2,1( ki .
Especially :,1k disassemble as ;2 qpxx
NMx
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Change the proper fraction to the sum of fractions – unknown coefficient( 待定系数法 )
653
2
xxx
)3)(2(3
xxx
,32
x
Bx
A
),2()3(3 xBxAx
),23()(3 BAxBAx
,3)23(
,1
BA
BA,
6
5
B
A
653
2
xx
x.
36
25
xx
Eg.1
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2)1(1xx
,1)1( 2
xC
xB
xA
21 ( 1) ( 1)A x Bx Cx x
Using the certain value to get CBA ,,
let ,0x 1 A let ,1x 1 B
let ,2x using A, B’s value 1 C
.1
1)1(
112
xxx2)1(
1
xx
Eg.2
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Eg.3
.1
51
52
2154
2x
x
x
)1)(21(1
2xx
),21)(()1(1 2 xCBxxA
,)2()2(1 2 ACxCBxBA
,1
,02
,02
CA
CB
BA
,51
,52
,54
CBA
,121 2x
CBxx
A
)1)(21(1
2xx
get
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Eg.4 Evaluate .)1(
12dx
xx
dxxx 2)1(
1dx
xxx
11
)1(11
2
dxx
dxx
dxx
1
1)1(
112
.1ln1
1ln Cx
xx
Sol.
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Eg.5 Evaluate
Sol.
.)1)(21(
12
dxxx
dxx
xdx
x
21
51
52
2154
dx
xx )1)(21(1
2
dxx
dxx
xxd
x
22 1
1
5
1
1
2
5
1)21(
21
1
5
2
.arctan5
1)1ln(
5
121ln
5
2 2 Cxxx
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Eg.5 Evaluate
Sol.
.
1
1
632
dx
eeexxx
let 6x
et ,ln6 tx ,6
dtt
dx
dx
eeexxx
6321
1dt
tttt6
11
23
dtttt
)1)(1(
16 2 dt
tt
tt
2133
136
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Ctttt arctan3)1ln(23
)1ln(3ln6 2
dtt
ttt
2133
136
.)arctan(3)1ln(23
)1ln(3 636 Ceeexxxx
23
)1ln(3ln6 tt dttt
td
22
2
11
31
)1(
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dxxx
x
32
52
)32ln(2
1 2 xx ;2
1arctan22 C
x
Eg.7
32
)32(
2
12
2
xx
xxd
dxx 2)1(
42
Eg.8
dxxx
x22 )32(
5
22
2
)32(
)32(
2
1
xx
xxd dx
x 22 ]2)1[(
4
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uduu
24
sec2sec4
4
1 2 tanLet x u
duu)2cos1(2
2
Cuuu )cossin(2
2
Cxx
xx
)32
)1(2
2
1(arctan
2
22
32
1
2
12
xx
原式
dx
x 22 ]2)1[(
4
udu2cos2
Cu
u )2
2sin(
2
2
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Eg.9 Evaluate
Sol. 原式 x
xd
14)1( 2 x )1( 2 x
21
1
d4x
x
2arctan
221 1
xx
21
221
ln21 xx
21 xxC
xx
x
x d1
21
2
2
12
1
x
xx
x d1
21
2
2
12
1
2)(21
21xx
)d( 1xx
2)(21
21xx
)d( 1xx
注意本题技巧
本题用常规方法解很繁
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Def.of trigonometric function rational expression:
2cos
2sin2sin
xxx
2sec
2tan2
2 x
x
,
2tan1
2tan2
2 x
x
,2
sin2
coscos 22 xxx
二、 The indefinite integral of trigonometric function rational expression
the finite number of operations of constantand trigonometric function,write as (sin ,cos )R x x
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2sec
2tan1
cos2
2
x
x
x
,
2tan1
2tan1
2
2
x
x
let2
tanx
u
,1
2sin 2u
ux
,
1
1cos 2
2
u
ux
ux arctan2
duu
dx 212
dxxxR )cos,(sin .1
211
,1
222
2
2 duuu
uuu
R
(万能置换公式)
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Eg.10 Evaluate .cossin1
sin
dxxx
x
Sol. ,1
2sin 2u
ux
2
2
11
cosuu
x
,1
22 du
udx
From the formula
dx
xxx
cossin1sin
duuu
u
)1)(1(
22
duuu
uuu
)1)(1(112
2
22
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duuu
uu
)1)(1()1()1(
2
22
duuu
211
duu
1
1
uarctan )1ln(21 2u Cu |1|ln
2tan
xu
2x
|2
sec|lnx
.|2
tan1|ln Cx
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Sol.( 二 ) dx
xx
x
cossin1
sin
dxxx
xxx
cossin2
]1)cos[(sinsin
dxxx )sec1(tan2
1
Cxxxx ]tanseclnsec[ln2
1
Cxx )]sin1ln([2
1
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Eg.11 Evaluate .sin
14 dx
x
Sol. (一) ,2
tanx
u ,1
2sin 2u
ux
,
12
2 duu
dx
dxx4sin
1du
uuuu
4
642
8331
Cu
uuu
]3
33
31
[81 3
3
.2
tan241
2tan
83
2tan8
3
2tan24
13
3 Cxx
xx
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Sol. (二)
dxx4sin
1dx
x
x 4
4
tan
sec
.cotcot31 3 Cxx
)(tantan
tan14
2
xdx
x
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Sol. (三)
dxx4sin
1dxxx )cot1(csc 22
xdxxxdx 222 csccotcsc )(cot xd
.cot31
cot 3 Cxx
Tips Comparing the upper methods, you’ll realize the formula method is not the optimization, using other method first, if you can not evaluate the indefinite integral, then think about the formula method.
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Eg.12 Evaluate .sin3sin
sin1
dx
xxx
Sol.2
cos2
sin2sinsinBABA
BA
dxxx
xsin3sin
sin1
dx
xxx
cos2sin2sin1
dxxx
x2cossin4
sin1
dxxx 2cossin
141
dxx2cos
141
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dxxx
xx2
22
cossincossin
41
dxx2cos
141
dxx
dxx
xsin
141
cossin
41
2 dxx2cos
141
xdxxdx
csc4
1)(cos
cos
1
4
12 xdx2sec
4
1
xcos41
xx cotcscln4
1 .tan
41
Cx
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Type: ),,( n baxxR ),,( n
ecxbax
xR
Methods: Get rid of the radical expression.
Eg.13 Evaluate
dxx
xx
11
Sol. Let tx
x
1,
1 2tx
x
三、 The indefinite integral of simple irrational function
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,1
12
t
x ,1
222
t
tdtdx
dxx
xx
11 dtt
ttt
22
2
1
21
1
2 2
2
tdtt
dtt
11
12 2C
t
tt
1
1ln2
.11
ln1
2
2
Cx
xx
x
x
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Eg.14 Evaluate .11
13
dxxx
Sol. Let 16 xt ,6 5 dxdtt
dx
xx 3 111
dtttt
523 6
1
dttt
16
3
Ctttt |1|ln6632 23
.)11ln(6131312 663 Cxxxx
Tips by using the least common multiple of the root exponent to get rid of the radical expression.
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Eg.15 Evaluate .1213
dxxx
x
Sol. First rationalize the denominator
dxxxxx
xxx)1213)(1213(
)1213(
dxxx )1213(
)13(1331
xdx )12(1221
xdx
.)12(31
)13(92 2
323
Cxx
.1213
dxxx
x
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1. 可积函数的特殊类型
有理函数分解
多项式及部分分式之和
三角函数有理式万能代换
简单无理函数
三角代换根式代换
2. 特殊类型的积分按上述方法虽然可以积出 ,但不
要注意综合使用基本积分法 ,简便计算 .一定简便 ,
内容小结
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一、 填空题:
1、
dxxx
CBx
x
Adx
x 111
323 ,其 A ____,
B ________ , C __________;
2、
dx
x
C
x
B
x
Adx
xx
x
11111
122
2
,
其中 A _____, B _____, C _______;
3、计算 ,
sin2 x
dx可用万能代换 xsin ___________,
dx _____________;
4、计算 ,
mbax
dx令 t ___, x ___, dx ____ .
练习题
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5、 有 理 函 数 的 原 函 数 都 是 _ _ _ _ _ _ _ _ _ .
二 、 求 下 列 不 定 积 分 :
1、 321 xxxxdx
; 2、 xxxdx
22 1;
3、 dx
x 411
; 4、 xdx
2sin3;
5、 5cossin2 xxdx
; 6、
dxxx
1111
;
7、
xdx
xx
11
; 8、 3 42 )1()1( xx
dx .
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三 、 求 下 列 不 定 积 分 ( 用 以 前 学 过 的 方 法 ):
1 、 dx
x
x31
; 2 、
dxxxx
sincos1
;
3 、 24 1 xx
dx; 4 、 dx
xx
3
2
cossin
;
5 、 dx
xx
28
3
)1(; 6 、 dx
xx
sin1sin
;
7 、 dx
xxxx
)( 3
3
; 8 、 dx
exe
x
x
2)1(;
9 、 dxxx 22 )]1[ln( ; 1 0 、 xdxx arcsin1 2 ;
1 1 、 dxxx
xx cossin
cossin; 1 2 、 ))(( xbax
dx.
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二 、 1 、 Cxx
x
3
4
)3)(1(
)2(ln
2
1;
2 、 Cxxx
x
arctan
2
1
)1()1(ln
4
122
4
;
3 、 )12arctan(4
2
12
12ln
8
22
2
xxx
xx
C )12arctan(4
2;
一、1、2,1,1; 2、-1,2
1,2
1;3、 221
2,
1
2
u
du
u
u
;
4、bax,a
bt2,dta
t2; 5、初等函数 .
练习题答案
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4、 Cx
3
tan2arctan
32
1;
5、 C
x
5
12
tan3arctan
5
1;
6、 Cxxx )11ln(414 ;
7、xx
xx
11
11ln C
x
x
1
1arctan2 , 或
Cxx
x
arcsin
11ln
2
;
8、 Cx
x
3
1
1
2
3.
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三 、 1 、 Cxx
1
1
)1(2
12 ;
2 、 Cxx )sinln( ;
3 、 Cx
x
x
x
2
3
32 1
3
)1(;
4 、 Cxxx
x )tanln(sec
2
1
cos2
sin2 ;
5 、 Cxx
x
4
8
4
arctan8
1
)1(8;
6 、 Cxx
2tan1
2, 或 Cxxx tansec ;
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7 、 Cx
x
66 )1(ln ;
8 、 Cee
xe xx
x
)1ln(1
;
9 、 Cxxxx
xxx
2)1ln(12
)]1[ln22
22
;
1 0 、 xxxx
arcsin124
)(arcsin 22
Cx
4
2
;
1 1 、 Cx
xxx
sin21
cos21ln
22
1)cos(sin
2
1;
1 2 、 Cxb
ax
arctan2 .