§

2

不定積分的幾種積分方法§

2

.

1§

2

.

2§

2

.

3湊微分法（第一換元法）變量代換法（第二換元法）分部積分法第一換元法第二換元法

f

[(x)](x)dx

f

(u)

du

[F

(u)

C]

u

(

x)dF[

(x)]

f

[(x)]

(x)dx設(shè)F(u)

f

(u),

u

(x)

可微,

則有

f

(u)du

f

[(x)]

(x)dx

d

F[(x)]

F[(x)]

Cu

(x)

[

f

(u)du]u

(

x)§

2.1

湊微分法(第一換元法)例1:求解:

設(shè)cos

udu

sin

u

C

2

1

sin(2x

1)

C且有積分公式

2

cos(2x

1)

1

d(2x

1)定理.

設(shè)即

f

[(x)]

(x)dx

f

[(x)]

d(x)湊微分u

(x)可微,則有換元公式

f

[(x)](x)dx

f

(u)du

F[(x)]

C,即

f

(u)du

F

(u)

Cd(2x

1)2

1

sin(2x

1)

C12例2例34

x2

d(4

x2

)x

d

e

x例4例5

dln

x

1

2

ln

x1

d(1

2

ln

x)2 1

2

ln

xdx

ex1

e

xex1

e2

x

dx

1

(ex

)2dexx

arctan(e

)

Cd(ax

b)a(1)

f

(ax

b)dx

1n(2)

f

(xn

)xn1

dx

1dcos

xx

f

(sin

x)cos

xdx

f

(cos

x)sin

xdx

dxnd(

x

)dsin

x(3)

f

(

x

)

1

dx

2(a

0)(7)

f

(ex

)ex

dx

dex(8)

f

(ln

x)1dx

dln

x(10)

f

(arcsin

x)

1

dx

f(arcsin

x)

d(arcsin

x)1

x2x(9)

f

(arctan

x)dx11

x2

f

(arctan

x)

d(arctan

x)(6)

f

(tan

x)

sec2

xdx

dtan

x例61a1

(

x

)2a2

1a1

(

x

)2dxa2

a

a

d(

x

)

aa

1

(

x

)2dxa1

(

x

)2

a

d

(

x

)(a

0)(a

0)例7dxsin

x

cos

x

cos

xdcos

xcos

x

dxsin

xsin

xdsin

x例8

C

C

1

ln

x

a2

a x

adx

dx

2a

x

a

x

a

1

2a

1 d(x

a)x

a2a

ln

x

a

d(x

a)x

a

C

1

ln

x

a2

a x

a1x2(x

a)(x

a)

a2

2a

1)1

12a x

a x

a

1

ln

x

a(x

a)

(x

a)

1

(

C

1

ln

a

x2

a

a

x例9dxsec2

x

sec

x

tan

xsec

x

tan

x

(sec

x

tan

x)sec

x

tan

x

d

(sec

x

tan

x)sec

x

tan

xcsc

xdx

ln

csc

x

cot

x或2

ln tan

x

C

C(P197

例4)12

2

2sin

x

cos

x

dx

2sec2

x2

tan

x22d(

tan

x

)tan

x2

dx

例10例111(x

1)2

( 2

)2d(x

1)2

12arctan

x

1

Cx8

2

d(x

2)8

2

(x4

)2d(x4

)例12例132

(

x

1)22(

5

)2

2

d

(x

1

)

d

e

x

arcsin

e

x

C1

(e

x)2例14(

a

b

).a2

sin2

x

b2

cos2

x

sin

x

cos

x

dxd(sin2

x)

2sin

x

cos

x

dxd(cos2

x)

2

cos

x

sin

x

dxa2

sin2

x

b2

cos2

xd(a2

sin2

x

b2

cos2

x)

1

解:原式=2(a2

b2

)

b2a2a2

sin2

x

b2

cos2

x

Cxx

Cdx

2第二換元法

f

[(x)](x)dxf

(u)

dut

1

(

x)f

[(t)](t)dt易求

第一換元法

第一換元法(湊微分法)解決的問題:

f

[(x)](x)dx

f

[(x)]d

(x)

[

f

(u)du]

u

(

x)難求

易求第二換元法(變量代換法)解決的問題:f

(x)d

x

x

(t)難求§

2

.

2變量代換法(第二換元法)例1:求2

2解:

令

x

a

sin

t,

t

(

,

)

,則dx

a

cos

t

d

t原式

a

cost

a

cost

d

t

a2

cos2

t

d

t22a)

Csin

2t

2(t

xata2

x2)

Ca2

x2aa22(1

cos

2t)d

t

x

ax

a2

x2

d

x

a2

arcsin

x

x

a2

x2

C2

a

2

2

(t

sin

t

cos

t)

C

2

(arcsin

aa2

a2定理.

設(shè)

x

(t)

嚴(yán)格單調(diào),可微,且變量代換即

f

[(t)]

(t)dx

F

(t)

C則有換元公式

f

(x)d

x

f

[(t)](t)dt

F[

1

(x)]

Ca2

x2

dxx

a

sin

taa

cos

t

(t

sin

t

cos

t)

C22a

cos

t

d

t

a2

x2

C

a2

arcsin

x

x2

a

2例22

2

a2

tan2t

a2x2

a2

a

sect解:

令

x

a

tan

t,

t

(

,

)

,

則dx

a

sec2

t

d

ta

sec2

t∴

原式

a

sectd

t

sec

t

d

t

ln sect

tan

t

C1xx2

a2ta

lnx2

a2a1(C

C

ln

a)

xa1

C例32

a2sec2t

a2x2

a2

a

tan

t則dx

a

sec

t

tan

t

d

t解:|

x

|

a,當(dāng)x

a時(shí),令x

a

sec

t,t

(0,

),∴原式d

ta

sect

tan

ta

tan

t

sec

t

d

t

ln

sec

t

tan

tx2

a2xt

C1

C1

(C

C1

ln

a)x2

a2ax

ln

a當(dāng)x

a

時(shí),

令

x

u

,

則u

a,

d

x

du

于是

u

2

a2d

u1

ln

u

u2

a2

C

x

x2

a2

C1

lna2

ln

C1x2x

a2(C

C1

2

ln

a)基本積分公式補(bǔ)充2*(25)

a

xa2

x2

C2

dx

a2

arcsinx

1

x2

a

21

x x2

a22ln

|

x

x2

a2

|

C22

a例4(2x)2

32ln 2x

124x2

9

C

2x

1

C

ln

x

1x2d(x

1)

2

1

d(2x)

1

d

u

ln

|

u

u2

a2

|

Cu2

a2(1)

f

(x,2

2令x

a

sin

t,t

(

,

)a2

x2

)

dx,令x

a

tan

t,t

(

,

)2

2令x

a

sec

t,t

(0,

)(

,

)

f

(x,

a2

x2

)

dx,

f

(x,

x2

a2

)

dx,2

2

f

(x,

n

ax

b

)

dx,令t

n

ax

bcx

d

)

dx,ax

b

f

(x,n§

3ax

bcx

d令t

n例5x

3

tan

t9

1

cos

t

d

t3xx2

9t

C

9

1

x9

x2例6x

t

63

2t

t6t

5dt)

dt11

t

6

(

t

2

t

1

6

1

t

3

1

t

2

t3

2

ln

1

t

C1

t3

11

6

t

dtt

3

1

(t

1)(t

2

t

1)(P.201.例15)例7(P.201.例16)解1：原式3

1

x

tdx

3t2

dt解2：原式1

x

tdx

dtP205

4-28,

10,

12,

16,

18,

26,

27,34,

36,

38,

39,

41,

42,

44,36.dx

1

ex(1

ex

)

ex或

1

e

x

dxe

x44.u

1

ex§

2.3

分部積分法u(

x)

v(x)

dx

u(

x)v(

x)

v(x)

u(x)

dx分部積分公式

u

dv

uv

v

du由導(dǎo)數(shù)公式

(u(x)v(

x))

u(x)v(

x)

u(x)v(x)積分得:u(

x)v(x)

u(x)v(x)dx

u

(x)v(x)

dx例1u

dv

u

v

v

du2另解:令u

cos

x

,dv

x

dx,則du

sin

xdx

,v

1x2x1

22

cos

x

x

(sin

x)

dx212更難積分!解:

令

u

x

,

dv

cos

x

dx,則du

dx

,v

sin

x∴原式

x

dsin

x

x

sin

x

sin

x

dx

x

sin

x

cos

x

C12∴

原式

cos

x

d(

2

x

)u(

x)

v(x)

dx

u(

x)v(

x)

v(x)

u(x)

dxu

dv

uvv

du選取u

及v(或dv)的原則:1)由v(或dv)容易求得v

;vu

dx比u

v

dx

容易計(jì)算.例2u

dv

uvv

du

x2

ln

x

dx

.解:

令u

ln

x

,

dv

x2

dx則

du

1

d

x,

v

1

x33x3

ln

x

1

x2

dx

9x3

ln

x

1

x3

Cx原式=

ln

x

d(3x

)

ln

x

x3

x3d(ln

x)例3u

dv

uvv

du解:

令

u

x2

,

dv

sin

xdx原式

x2d(

cos

x)

2

x

d

sin

x(x

sin

x

sin

xdx)例4u

dv

uvv

du

x2

ex

dx

.

x2

d

ex

x2ex

ex

dx2

x2ex

2

ex

x

dx

x2ex

2

xd

ex

x2ex

2(xex

ex

d

x)

x2ex

2(xex

ex

)

C

ex

(x2

2x

2)

Cdv

ex

dx例5u

dv

uvv

du

1

x211

x2x2

11

dx))

dx

]2

x

arctan

x

dx.

u

arctan

x12arctan