微积分-2分部积分法

§

2.3

分部积分法u(

x)

v(x)

dx

u(

x)v(

x)

v(x)

u(x)

dx分部积分公式

u

dv

uv

v

du由导数公式

(u(x)v(

x))

u(x)v(

x)

u(x)v(x)积分得: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,则du

sin

xdx

,v

1x2x1

22

cos

x

x

(sin

x)

dx212更难积分!解:

令

u

x

,

dv

cos

x

dx,则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

)分部积分公式u(

x)

v(x)

dx

u(

x)v(

x)

v(x)

u(x)

dxu

dv

uvv

du选取u

及v(或dv)的原则:1)由v(或dv)容易求得v

;vu

dx比u

v

dx

容易计算

.例2u

dv

uvv

du

x2

ln

x

dx

.解:

令u

ln

x

,

dv

x2

dx则

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

x

d(x

)2

1

(x2arctan

x

x2

darctan

x)2

12

(x

arctan

x

122[x

arctan

x

(12

1

[x2arctan

x

(x

arctan

x)]

C2

1

[(x2

1)arctan

x

x]

C(1)dx,P

(x)ea

xna

xn

v令

P

(x)

u,

e

Pn(x)

sin(ax

b)

dx,

Pn

(x)

cos(ax

b)dx,(Pn

(x):n次多项式)

f

(x)

ln

x

dx,

f

(x)

arctan

x

dx,

f

(x)

arcsin

x

dx,令Pn

(x)

u,sin(ax

b)

v令Pn(x)

u,cos(ax

b)

v(n次分部积分)令

ln

x

u,

f

(x)

v令

arctan

x

u,

f

(x)

v令

arcsin

x

u,

f

(x)

v例6u

dv

uvv

du

x

arccos

x2

121

x2d(1x

)2

x

arccos

x

1

2 1

x2

C

x

arccos

x

x

arccos

x

x

dx1

x2dv

dx,

xdarccos

x

x

arccos

x

1

x2

C例7u

dv

uvv

du

e3

x

cos

2x

dx

.e

cos

2x1

3x3e

dcos

2x

133xe

dsin

2x]3x3x3xcos

2x

d

e解:

e3

x

cos

2xdx

(u

cos

2x

,

v

e3

x

)e

cos

2x313

13xe

sin

2x

d

x3x3xsin

2x

d

e[e

sin

2x

3x(u

sin

2x

,

v

e

)232929e

cos

2x

d

x943x93

xe

(3cos

2x

2

sin

2x)13故

原式

1

e3

x

(3cos

2x

2

sin

2x)

C

1

e3x

cos

2xe

cos

2x

31313x循环积分ea

x

ea

x

cos

bx

dx

2a

b2

(b

sin

bx

a

cos

bx)

Cea

x

ea

x

sin

bx

dx

(a

sin

bx

b

cos

bx)

C

b23xe

cos

2x

dx

e

(3cos

2x

2

sin

2x)

Ca2

1

133x例8.

e3

x

cos2

x

dx123

x112

23x3

x

e

dx

e

cos

2x

dxe

(1

cos

2x)

dx

1

e3x

1

e3

x

(3cos

2x

2sin

2x)

C6

26内容小结(代换:x

(t))第二类换元法3.分部积分法u

vdx

u

v

uv

dx

ud

v

uv

v

d

u求不定积分的基本方法直接积分法利用基本积分公式和运算法则，求不定积分换元积分法

第一类换元法

(凑微分:

d[(t)]

)例9u

dv

uv

v

du

tan

x

ln

cos

x

tan

x

sin

x

dxcos

x

tan

x

ln

cos

x

(sec2

x

1)

dx

tan

x

ln

cos

x

tan

x

x

Cdv

dx

cos2

x

ln

cos

x

d

tan

x

tan

x

dln

cos

x例10u

dv

uvv

du1

x2

1 ln(x

1

x2

)

d

1

x22

ln(x

(P.203.例20)1

x2

)2d(1

x

)例11解:

令x

(

2

e x

1)

Cx

t

,

则

x

t

2

,

dx

2t

d

t

2t

e

t

d

t

2

t

d

e

t

2(tet

e

t

d

t

)

2

(t

e

t

et

)

Cu

dv

uvv

du例12解:

sec3

x

dx

sec

xd

tan

x

sec

x

tan

x

tan

x

d

sec

x

sec

x

tan

x

tan2

x

sec

xd

x

sec

x

tan

x

(sec2

x

1)

sec

xd

x

sec

x

tan

x

sec3

x

d

x

sec

xd

x2∴原式

1

sec

x

tan

x

ln

sec

x

tan

x

Csec3

x

dx

.

sec

xd

x

ln

|

sec

x

tan

x

|

C求例13解1:x

a

tan

t,dx

a

sec2tdtx2

a2a2xta22a

xa

a2

x

sect

tan

t

ln

sec

t

tan

t

C2a2

x22

a2

x

a

Ca2

x2

a

2

lndx

a

sec

t

a

sec2tdt

a2

sec3

tdt(变量代换)x2

1

x

a222a

a2

ln|

x

x2

2a

|

C2

ln

a)(C

C1

a2a2x2

a2

x

x

dx

x x2

a2x2dx

a2x2

(x

a

)

a2

22x2

a2

dx

x

x x2

a2

x2

a2

dxx2

a2a2

d

x

2x2∴

原式=

1

x

a22

a2

ln|

x

x2

a2

|

C解2:

(分部积分)例13

x22x2

a2

|

Cx2

a2

a2

ln|

x

例14求已知

的一个原函数是

x

f

(x)

dx

x

d

f

(x)

x

f

(x)

f

(x)

dx说明:

此题若先求出x，求积分反而复杂.x

f(x)

dx

xx2(

cos

x

2sin

x

2

cos

x

)

d

xx2x

x解:

f

(x)

(cos

x

)

sin

x

cos

xx2x

x

x

(

sin

x

cos

x

)

cos

x

C

sinx

2

cos

x

Cxf

(x)

dx＝cos

x

C例15解1:

原式

x

a

tan

ta

sec2t

dt2xx2

a2ta

1

(t

1

sin

2t)

C

1

(t

sin

t

cos

t)

C2a3

2a3例15解2:

原式dxx2

(x2

a2

)2

a2

x2)

a2x22

2x

a2d

xx2x

d(1

x

12a2

a2

2a

a2

)21x

d(x2

a2

)(x2u

dv

uvv

du例15(P.204.例23)解:

In得递推公式n1

1

Ia21

a2

)n1(x2x

d例15(P.204.例23)递推公式n1nI

2n

32(n

1)a2

a2

)n1x2a2

(n

1)(x2I

n

2,

3,

11I1

1

arctan

x

Ca

a1

x2a2

(x2

a2

)

2a2

I arctan

x

Ca12a31x2a2

(x2

a2

)M

x

Ndx

(x2

px

q)n例16dxx

2x

3

212

x

2x

3

1

ln(x22

2x

3)

3

arctan

x

1

C2

2221

(2x

2)

33d(x

1)(x

1)2

(

2)2

2d(x

2x

3)例16dx(x

2x

3)

22

21

(2x

2)

3

1 d(x

2

2x

3)2

(x

2

2x

3)2[(x

1)2

(

2)2

]2

3

d(x

1)

arctan

x

1

C

(u2

a2)2duarctan

u

Ca12a31

u2a2

(u2

a2)4

x2

2x

3

4

2

21

3

x

1

32(x2

2x

3)M

x

Ndx

(x2

px

q)n作业P206

4－245,

46,

49,

50,

51,

52,

5357,58,

60,

61,

62u

ex

145.49.x

e

dx23

x

1222

xx

d(e

)§

3

特殊函数的不定积分§

3.1

有理函数的不定积分§

3.2

三角函数有理式的不定积分§

3.3

某些无理函数的不定积分§

3.1

有理函数的不定积分

(x

a)nn

1nI

dx

(x2

a2)nI1

1

arctan

x

C,a

a递推公式n

2,

3,1

n

1

(x

a)1n

C,

n

2,3,

dx

ln x

a

C,M

x

NAd

xd

x,

(x

a)k

(x2

p

x

q)k(k

N

,

p2

4q

0)有理函数R(x)

Pn

(x)

a0

xn

a1

xn1

anQm

(x)为假分式;

m

n时,为真分式m

n

时,有理函数相除多项式+真分式分解A,(x2

p

x

q)kM

x

N(x

a)k(

k

N,

p2

4q

0

)若分分式

之和有理函数的不定积分A M

x

Nd

xd

x,

(x

a)k

(x2

p

x

q)k(k

N

,

p2

4q

0),du

du

(u

a)k

(u2

a2)k一切有理函数的原函数一定是初等函数原函数的类型：多项式、有理函数、对数函数及反正切函数例1

将下列真分式分解为部分分式

:解:

(1)用拼凑法

(

P.213.例8

)1x2

(1

x2

)2

x2

(1

x2

)21x2

(1

x2

)11(1

x2

)2(1

x2

)

x2(1

x2

)

x2

x2

(1

x2

)(1

x2

)2x2

1

11

x21(1

x2

)2(2)赋值法(P.209.例3)55

2A

D

262A

4C

4D

5令x

1

可得B

1

,分别令x

0,1,2

代入等式两端x2A

B

Cx

D(x

1) (x

1)2

2x

2A(x

1)(x2

2x

2)

B(x2

2x

2)

(Cx

D)(x

1)2

x10

A

2C

D

015(x

1)2125(x

1)原式C

1

,

D

825

25x

825(x2

2x

2)A

1

,

B

125

5例2 arctan

x

Ca12a31x2a2

(x2

a2

)50

1x2

2x

221

(2x

2)

725

(x

1)2

1d(x2

2x

2)

7

d(x

1)例2例32

(x

2x

2)

(2x

2)

dx(x2

2x

2)2d(x

1)

(x

1)2

1

(x2

2x

2)2d(x2

2x

2)

arctan(x

1)

1x2

2x

2

C例4

x41dxdx(x2

1)

(x2

1)(x2

1)(x2

1)12

]dx1

1[2

1x

1

x2

12

2

2u

adu

C

1

ln

u

a2a u

a2

1

arctan

x

C(P.213.例7

)12

2

x

1

x

1x2或

1

[1

(

1

1 )

1

]dx例5x

11)x4

1dx2x2x2x2

11

1dx

x(x

1

)2

2dx

x

d(x