不定積分基本積分法

6.1

不定積分的基本積分法第6章

積分法性質(zhì)1

[f

(x)

g(x)]dx

f(x)dx

g(x)dx;證

f

(x)dx

g(x)dx

f

(x)dx

g(x)dx

f(x)

g(x).等式成立.（此性質(zhì)可推廣到有限多個(gè)函數(shù)之和的情況）6.1.1

不定積分的性質(zhì)

kf

(

x)dx

k

f

(

x)dx.性質(zhì)2 設(shè)

k

是常數(shù),

(k

0)

則由此可知:

不定積分運(yùn)算是線性運(yùn)算,

即

[k1f(x)

k

2g(x)]

dx

k1

f(x)dx

k

2

g(x)dx解:)dx1

x23

(1

x2dx1

x2dx

21

x2

311

3arctan

x

2arcsin

x

C例1

求積分3

21

x2

)dx.

(1

x22例2求積分

x(1

x2

)dx.1

x

x21

x

x2

x

(1

x2

)解:

x(1

x2

)dx

x(1

x2

)

dxx

1

x12

1

dx

x1dxdx

1

1

x2

arctan

x

ln

x

C.例3、計(jì)算dx.x22

x2

x2

x

1

2

x

x

dxdx

2x22

x2

x2

x

1解：xdx

x2dx

2

dx

C123

2

x

3x

2x例4

計(jì)算2

sin2

xdx.解：利用三角恒等式x

1

1

cos

x2

2sin2

sin2

xdx

1

(1

cos

x)dx2

2

1

dx

1

cos

xdx2

2

1

x

1

sin

x

C2

2例5.

計(jì)算

cos

dx.sin2

解：將被積函數(shù)變形sin

sin

cos

csccot

sin2

cos

1

cos

dx

csc

cot

d

csc

Csin2

例6.

計(jì)算

a

x2dx.解：

ax2dx

a2

axdx

a2

axdxln

a

ln

a2

a

1a

C

a

Cx

x

2例7.求積分解:dx.1

cos

2

x11

cos

2

x1dx

dx1

2cos2

x

11

1

12 cos2

x2dx

1

tan

x

C

.說(shuō)明：以上幾例中的被積函數(shù)都需要進(jìn)行恒等變形，才能使用基本積分表.例8.

計(jì)算

1

x2

dx.x4解：

1

x2

dx

1

x2x4

1

1dxx4例9.

計(jì)算

dx.sin2

x

cos2

x解：1 cos2

x

sin2

x

sin2

x

cos2

x

dx

sin2

x

cos2

x

dx1

1

(sin2

x

cos2

x)dx

tan

x

cot

x

C1

x2(x

2

1)(

x2

1)

1dx

(x2

1

11

x21)dx

1

x3

x

arctan

x

C3問(wèn)題:怎樣求

cos2

x

sin

x

dx

(x

1)99

dx

等6.1.2.

不定積分的換元法A.第一換元法（“湊”微分法）定理1(不定積分的第一換元積分法)若已知

f

(x)dx

F

(x)

C而u

(x)是任一可微函數(shù)則

f

(

x)

'(

x)

dx

F

(

x)

C

f

(

x)

'(

x)

dx

f

(

x)

d(

x)(令u

(x))

f

(u)

duF

(u)

f

(u)

F

(u)

C

F

(

x)

C回代證明:(步驟)第一步:湊微分第二步:作變換第三步:求積分第四步:回代例1.

e

2

x

dx22

x

e

d2

x(令u

2

x)22

C2u

u

2

x

e

du

e

C

e回代例2.

1

dx1

x1

d(x

1)

(令u

x

1)1

x

u

1

du

ln

|

u

|

C

ln

|

x

1

|

C3

x

1dx

1例3.a則

f

(ax

b)

dx

1

F

(ax

b)

C13

x

1

3

1

d(3

x

1)3

1

ln

|

3

x

1

|

C一般地

若已知

f

(

x)

dx

F

(

x)

C3

5例4.

(3x

2)4

dx

1

1

(3

x

2)5

Cx

3

99

(

50

)dx

50

100

501

x

3

100

C2x

2

x

e

dx1

(

x)dx

ex

221

2

221

x

e

d1

x

2

e

Cx

221例5.

dx

15

16x

4x2

dx

1

(2

x

4)21

d(2

x

4)

2

1

(2

x

4)22

1

arcsin(

2

x

4)

C例6.例7.

e

xex

dxx

x

e

x

e

e

dx

e

e

d(ex

)x

ee

C

sec4

x

tan3

x

dx

sec2

x

tan3

x

sec2

x

dx

(1

tan2

x)

tan3

x

dtanx

(tan3

x

tan5

x)

dtanx

1

tan4

x

1

tan6

x

C4

6(或

sec3

x

tan2

x

d

sec

x)例8.secm

x

tan2n1

x

dx

secm1

x

tan2n

x

dsecx

secm1

x

(sec2

x

1)n

dsecx

sec2m

x

tann

x

dx

sec2m2

x

tann

x

dtanx

(tan2

x

1)m1

tann

x

dtanx一般地,例9.

tan

x

dxcos

xsin

x

dx

cos

xdcos

x

ln

|

cos

x

|

C試求：

cot

xdx

cos

x

dx

dsin

x

ln

|

sin

x

|

Csin

x

sin

x例10.

sec

x

dx

dx

cos

x1cos

x

dxcos

x2dsin

x21

sin

x(1

sin

x)(1

sin

x)dsin

x

dsin

x1

sin

x

12

1

sin

x1

12

1

ln(1

sin

x)

ln(1

sin

x)

C

ln

C

ln2 1

sin

x

21 1

sin

x

1

(1

sin

x)2

Ccos2

x

ln

|

sec

x

tan

x

|

C例10.解二:

sec

x

dx

ln

|

sec

x

tan

x

|

C試求：

csc

x

dx

ln

|

csc

x

cot

x

|

C

dxsec

x(sec

x

tan

x)sec

x

tan

xd(sec

x

tan

x)sec

x

tan

x例11.

cos(3

x

8)

cos(8x

3)

dx

cos(11x

5)

cos(5x

11)

dx12

2

11

5

1

1

sin(11x

5)

1

sin(5

x

11)

C2210

1

sin(11x

5)

1

sin(5

x

11)

C類似的,sin

x

cos

x

dx

sin

x

sin

x

dx先”積化和差”例12.sin2

x

dx

dx1

cos

2x224

1

x

1

sin

2x

C類似的,

sin2n

x

dx

或

cos2n

x

dx(半角公式降次)dx.

a21

x2a21

1

1

a

d

x

a

a2

x

2

a

21

x

dx

11a

1

arctan

x

C

.例13.

Cx3

3adx

1

arctan1

3

x2a2

x2a試求：

1

dx

arcsin

x

C例如:dx.1

x2

8

x

25dx

1(

x

4)2

9

1

arctan

x

4

C

.3

3例14.(

dx

1

arctan

u

C

)u2

a

2

a

a(u

x

4)dx.x

a2

x2

1

122d

(

x2

)a

x21

12d

(

x2

a2

)2

a

x22

1

ln(

x2

a2

)

C例15.

dx4

x

12

x

134

x

12x

5

dx124

x

12x

1314

x

2

12x

1322

1 d(4

x

12x

13)

dx

52(2

x

3)

42

1

ln(4x2

12x

13)

d2x

312x

3

2

45224

2

1

ln(4x2

12x

13)

5

arctan

2x

3

C例16.(分子一次湊分母導(dǎo)數(shù))1dx.

x2

a2(

x

a)(

x

a)

1

)dx12a x

a x

adx

1

(1dx

1

1x

adx

2a

x

a12a

1

(ln

x

a

ln

x

a

)

C

1

ln

x

a

C2a x

a例17.1

(

x

a)(

x

b)dx.

1

(

1

1

)dxa

b x

a x

bdx

dx

a

b

1

1

1x

a x

b(ln

x

a

ln

x

b

)

Ca

b1一般地,

dxx

4

x

5x

72

dx

92

x

42

12x

4

x

5

dx12x

4

x

5x

2

4

x

5d(

x

2

4x

5)121

9

(

x

5)(

x

1)

dx1

1

15

x

1

dx2

6

x

1

ln

|

x

2

4x

5

|

9

1

ln

|

x

2

4x

5

|

3

ln

x

5

C2

2

x

1例18.(分子一次湊分母導(dǎo)數(shù))x3

x

ln(arcsin

3

x

)

dxx1

9

arcsin

3ln(arcsin

3

x

)ln

31x

d(3

)x1

(3

x

)2

arcsin

3arcsin

3

xln(arcsin

3

x

)1x

d(arcsin3

)ln

31

ln(arcsin

3

x

)

dln(arcsin3

x

)ln

31ln(arcsin

3

x

)2

C2

ln

3例19.1

f6

(

x)2(

x)

f

(

x)

dx例20.

f

df

(

x)f

2(

x)61

f

(

x)1

1

df

3

(

x)1

f

6

(

x)33

1

arcsin

f

3

(

x)

C例21.

f

(

x)f

[

f

(

x)

1]

1

dx

f

[

f

(

x)

1]

1

df

(

x)

f

[

f

(

x)

1]

df

(

x)

df

(

x)

f

[

f

(

x)

1]

f

(

x)

C定理2(第二類換元法)若x

(t

)單調(diào)、可導(dǎo)，'(t

)

0且

F'(t

)

f

(t

)

'(t

)

f

(

x)

dx

f

(t)

(t

)

d

t

F[

1

(

x)]

C則

f

(t)(t

)

d

t

F

(t

)

C

F[

1

(

x)]

C證明：

f

(

x)

dx

令

x

(t

)

f

(t

)

d

(t

)B.

不定積分的第二類換元法步驟:作變換積分回代a

2

x

2

dx

(a

0)例1.解：令x

a

sin

ta2a

2

x2

dx

a

cos

t

a

cos

t

d

t

a

2

1

cos2t

d

t21x2

2xsin

t

aata

x2a

2aa

2

x

2cos

t

x)

Cx a

2

x

2(arcsin

a

aa22a

2

arcsinx

xa

2

a

2

x2

Cdx

a

costdt(

t

)2

2a2

(1

sin2

t

)

a

cost,則

a2

x2

(t

sin

2t

)

C

2(t

sin

t

cost

)

C2a21.三角代換(目的:去根號(hào))(a

0)

dtdx a

sec2

ta

2

x2

a2

a

2

tan2

t

dt

sec

t

dta

sec

ta

sec2

t

1

ln

|

sec

t

tan

t

|

Ctx2

2a

xatan

t

xasec

t

22aa

x

lnaa

2

x

2

x

C1

ln

|

x

aa2

x2

|

C

t

)，則2

2a2

x2解：令

x

a

tan

t

(例2.

dx

a2

x2

a2

(1

tan2

t

)

a

sec

t,

dx

a

sec2

tdt

ln

|

x

x

2

a

2

|

Cx

2

a

2dxxxa

2222

2xa

x

dx

a

x

arcsin

C2

2

ax

a

dx

2x

a

ln

|

x

x

2

a

2

|

C2a

22

222若被積函數(shù)中含有x

a

sin

t

或x

a

cos

ta

2

x

2a

2

x

2x

2

a

2例如：作變x

a

tan

t

或x

a

cot

t換x

a

sec

t

或x

a

csc

t總結(jié):3

x

19

x

2

6

x

8dx

d(3x

1)3

x

1(3

x

1)2

9(令3x

1

3sec

t)3

tan

t1313

3sec

t

3sec

t

tan

t

d

t

tan2

t

dt

(sec2

t

1)

d

t

tan

t

t

C9x26x

839

x

2

6

x

8

3

arccos

C3

x

13

x

13x

1t3tan

t

3cos

t

39

x

2

6

x

8例3.dx例4.7x(1

x

)t(令x

1)(1

1)117tt)

d

t

t1

(2

d

t1

tt

67771

t7

1

1

d

t

7

1

ln

|

1

t

7

|

C7

1

ln

|

1

1

|

Cx7

x(1

xn

)dx一般nxn

1

ln

|

1

1

|

C2.倒數(shù)代換

dxx2

1

x21令x

t21

(1)2(

)

1

1t

t)

dt

1t

2

(

d

tt

2

1|

t

|

t

2

12

t

d

t

1

1

d(t

2

1)t

2

1t

2

1

C

(

1

)2

1

Cx

Cx1

x

2

例5.e

x

1

dx例6.e

x令

t

1則x

ln(t

2

1)原積分

t

2tt

2

1)d

t1

t

2

d

t

2(1

1

2t

2arctan

t

Ce

x

1

Ce

x

2

1

2arctan3.令整個(gè)根號(hào)為新變量解:例7.1

1

3

x

2

dx1

tt

1

1dt

32)dt

1

1

t

3 (t

1

2

3

(t

1)2

3

ln

|

t

1

|

C2

3

(3

x

2

1)2

3

ln

|

3

x

2

1

|

C令t

3

x

2

3t

dt1

t2說(shuō)明當(dāng)被積函數(shù)含有兩種或兩種以上的根式kx,,

lx

時(shí)，可采用令x

t

n（其中n為各根指數(shù)的最小公倍數(shù)）例8.求dx.1x(1

3

x

)解:

令

x

t

6

dx

6t

5dt,dxx(1

3

x

)1

6t

5

6t

2t

3

(1

t

2

)

dt

1

t

2

dt1

6

(1

1

t

2

)dtx

6arctan

6

x

C

6t

6arctan

t

C

66乘積求微分公式d(uv)

u

dv

v

du兩邊積分uv

u

dv

v

du改寫為

u

dv

uv

v

du分部積分公式例1

x

cos

x

dx則duv

s注意(I)v

要容易求得；(II)u'要比u

更簡(jiǎn)單解:

令

u

dv

co

x

cos

x

dx

x

sin

x

sin

x

dx

x

sin

x

cos

x

C6.1.3

不定積分的分部積分法例

2

arctan

x

dx則du

1

x

2dx令u

arctan

xv

x

x

arctan

x

dx1

xx2

x

arctan

x12ln(1

x

)

C2例

3

x2e

x

dx

x

2de

x

x2e

x

(

e

x

2

x

dx)

x2e

x

2

x

de

x

x2e

x

2xe

x

2

e

x

dx

x2e

x

2xe

x

2e

x

C被積函數(shù)多項(xiàng)式

正、余弦函數(shù)多項(xiàng)式

指數(shù)函數(shù)多項(xiàng)式反三角函數(shù)多項(xiàng)式

對(duì)數(shù)函數(shù)畫紅線者拖到d后面ex

sin

x

或ex

cos

x兩者都可

3x

2

ln(

x

1)

dx

ln(

x

1)

d(

x

3

1)例43x

3

1

(

x

1)

ln(

x

1)

x

1

dx

(

x3

1)

ln(

x

1)

(

x

2

x

1)

dx

(

x3

1)

ln(

x

1)

1

x

3

1

x

2

x

C3

2x

e

dx例535

x31333

xx

e

d

x

3133

xx

d

e33

x13x

e

313x

3e

d

x

3

xx

e

13

33

1

e

x

3

C

sec3

x

dx

sec

x

dtan

x例6

sec

x

tan

x

tan2

x

sec

x

dx

sec

x

tan

x

(sec2

x

1)

sec

x

dx

sec

x

tan

x

sec3

x

dx

sec

x

dx2

1[sec

x

tan

x

ln

|

sec

x

tan

x

|]

C例7.求積分

ex

sin

xdx.解

ex

sin

xdx

sin

xdex

ex

sin

x

exd

(sin

x)

ex

sin

x

excos

xdx

ex

sin

x

cos

xdex

ex

sin

x

(ex

cos

x

exd

cos

x)

ex

(sin

x

cos

x)

ex

sin

xdx2x

exsin

xdx

e

(sin

x

cos

x)

C

.注意循環(huán)形式例

8.

e

x

dx解:

令

t

則

x

t

2d

x

2t

d

t

2t

et

2

et

d

t

2t

et

2et

Cx

x

2 x

e

2e

Cx

dx

et

2t

dt

2t

det

e2例

9.已知

f

(

x)的一個(gè)原函數(shù)是e

x

,

求

xf

(

x)dx.解:

xf

(

x)dx

xdf

(

x)

xf

(

x)

f

(

x)dx,

f

(

x)dx

f

(

x),

f

(

x)dx

e

x2

C,2上式兩邊同時(shí)對(duì)x求導(dǎo),

得

f

(

x)

2

xe

x

,

xf

(

x)dx

xf

(

x)

f

(

x)dx2

2

2x2e

x

e

x

C

.Q(

x)P(

x)R(

x)

xmma

a

x

a

xn

0

1

n

b

b

x

b0

1n

m

時(shí)R(x)是真分式n

m

時(shí)R(x)是假分式任一假分式可以通過(guò)多項(xiàng)式除法化為一個(gè)多項(xiàng)式與一真分式之和．x

2

2

x

1x4

5例如2

4

x

2

x

2

x

3

x

2

2

x

16.1.4

幾種特殊類型函數(shù)的積分1.有理函數(shù)的積分法(I)有理函數(shù)具體可表示為Q(

x)(II)

將真分式R(x)

P(x)分解為部分分式．原理Q(x)在實(shí)數(shù)范圍內(nèi)可分解為一次因式與二次因式之積x

3

x

2

2

x(i)

2

x

3

2

x

3x(

x

2)(

x

1)B

A

x x

2

x

1Cx(

x

1)3x

3

1(ii)

A

B

C

(

x

1)3

(

x

1)2x

x

1D(

x

1)(

x

2

x

3) x

4

A

Bx

C

(iii)x

1x

2

x

3(

x

2)(

x

2

1)22

x

2

2x

13(iv)A

x

2

(

x

2

1)2Bx

Cx

2

1Dx

E項(xiàng)。形如1(g(a)

0)拆開(kāi)必含(

x

a)g(a)1(

x

a)g(

x)1(

x

5)

3(

x

2)

(3)1(

x

2)(

x

5)1例如：確定待定系數(shù)(III)通分后，比較同次冪系數(shù)四種部分分式的積分類型(IV)

dx

x

adx(i)(

x

a)n(ii)

dxx

2

px

qAx

BAx

B(iii)

dx(

x

2

px

q)m(iv)(

p2

4q)(

p2

4q)

ln

|

x

a

|

C

C

n11

n

(

x

a)1

1

dxx

2

2

x

43

x

2例1

dx

5x

2

x

4

3 2

x

2

22

dxx

2

2

x

41x

2

2

x

423 d(

x

2

2

x

4)

5

(

x

1)2

3d(

x

1)

3

ln(

x

2

2x

4)

5

arctan

x

1

C2

3

3

dx(

x

2

4x

13)28

x

31例2

d

x

152

x

4

4

(

x

2

4

x

13)2(

x

2

4

x

13)2d

xd

x

4

(

x

2

4

x

13)2d(

x

2

4

x

13)[(

x

2)2

9]2

15x

2

4

x

13

4

15d

x[(

x

2)2

9]2

dx(

x

2

4x

13)28

x

31例2

4x

2

4

x

13

15d

x[(

x

2)2

9]22[(

x

2)

9]d

x(令x

2

3tan

t

)2

81sec4

t

d

t3sec2

t1272cos

t

d

t

d

t1 1

cos

2t22754t

1

sin

2t

C1083x

23x2

4x

13t3tan

t

x

254

3x

2

4

x

13x

2x

2x

2

4

x

13

C54

1

arctan

x

2

118

x

2

4

x

13

C54

3

1

arctan

x

2

1x

2

4

x

1354

3

15

arctan

x

2

1518

x

2

4

x

13x

2

C

dx(

x

2

4x

13)28

x

31例2

4x

2

4

x

13

4

15[(

x

2)2

9]2d

xx

2例

3

1

x4

d

xx

2

(1

x

2

)(1

x

2

)

d

x2

2

1

1

1

x

1

x2

1d

x21

arctan

x

d

x4

1

x

1

x

1

1

1

24

1

arctan

x

1[ln

|

1

x

|

ln

|

1

x

|]

CR(sin

x,cos

x)

d

x(I)(

x

)x2t

tan萬(wàn)能代換：則

x

2d

t1

t

2d

x

2sin

x

2sin

cos2

2x x

2tan

cos2x

2

x

22

tan22

x2x1

tan21

t2t2xcos

x

2

cos2

1

1

1

t

2221

t

21

t2

,

2

)

2

d

t2t

1

t

2

21

t

1

t

1

tR(sin

x,cos

x)

d

x

R(

2.三角有理函數(shù)的積分法d

x4

5cos

x1例42(令t

tan

x

)4

51

t

21

t

2d

t

1

t

22

d

t9

t221

1

1

d

t3

3

t

3

t

Cln1 3

t3 3

t23

tan

x2

C3

tan

x3

1

ln數(shù)，即R(

sin

x,cos

x)

R(sin

x,cos

x)(則作代換：若R(sin

x,cos

x)是cos

x

的奇函數(shù)即R(sin

x,

cos

x)

R(sin

x,cos

x)則作代換：t

sin

xsin4

x例5.

tan

x

cos6

x

d

x

d

x

(令t

sin

x

則cos

x

dx

d

t

)sin

x3cos5

x

t

3(1

t

2

)2

d

t

(

t

)

d

tt1

2t32

1

2ln

|

t

|

1

t

2

C2

2

ln

|

sin

x

|

1

sin2

x

C2sin2

x2t

21則可作代換：t

tan

xcos4

xsin2

x

1例6

d

x

cos2

x

cos2

xsin2

x

1 d

x

(tan2

x

sec2

x)

dtan

x

(2

tan2

x

1)

dtan

x3

2

tan3

x

tan

x

C若R(

sin

x,

cos(III)ax2

bx

c

)已解決的無(wú)理函數(shù)：R(x,n(I)

R(

x,

d

x例732(

x

1)

(

x

1)ax

b)

dxx

1

16(令

t

x

1)

t

6

t

4t

3

1

6

t

5d

t

6

d

tt(t

3

1)2t