高數上4四章4節(jié)

§4.4

幾種特殊函數的不定積分一、有理函數的不定積分：1、任意整式h(x)都可分解為若干個一次式和判別式為負的二次式之積.例、x4

+4

=(x4

+4x2

+4)-4x2=(x2

+2)2

-(2x)2=(x2

+2x+2)(x2

-2x+2)nP

(x)Qm

(x)n

<

m.2、稱為真分式Q(x)3、任意有理式R(x)可分成整式h(x)與真分式P(x)之和.x4

-

x3

+

x2

+

2x2

+

x

-1例、R(x)==(x2

-2x

+4)

+

-6x

+6x2

+

x

-1(其中p2

-4q

<0,,r

2

-4s

<0)Q(x)則：真分式P(x)=0(

x2

+

px

+

q)l

(

x2

+

rx

+

s)m若Q(

x)

=

b

(

x

-

a)a

(

x

-

b)b4、真分式的性質：+++(

x2

+

px+q)lMlx+Nlx2

+

px+q+

M1x+N1++

++(x-a)aAa(

x-a)2A2A1(x-a)Rm

x+Smx2

+rx+s

(

x2

+rx+s)mR1x+S1+++BbB2(

x

-b

)b(

x

-b

)2B1(

x

-b

)+++

+Q(

x)

P(

x)

=(其中Ai，，Bi；Mi、Ni，，Ri、Si是常數)注1、此稱部分分式法：將真分式寫成基本分式之和.2、要將Q(x)分解到不能再分為止.3、分法：(1)確定基本分式的個數及形式；

(2)確定常數.4、確定常數：待定系數法；特值法.x2

-5x+6例1、將下列各式寫成基本分式之和：1(3)x(x-1)x+1(1)

x+3

=A +

Bx-2

x-3(2)

x

+2

=

A

+(x

-1)2

x-1

(x-1)22A

B

C2

=

+

+x

x-1

(x-1)A

Bx

+C(4)

=

+x(x2

+1)

x x2

+1=

A(x-3)+B(x-2)(x-2)(x-3)B

=

A(x-1)+B(x-1)2=

A(x-3)+B(x-2)(x-2)(x-3)

A(x

-3)

+B(x

-2)

”

x

+3(*)在(*)中分別令x

=2,3得：

B

=6-

A

=5

A

=-5,

B

=6=

A

+

Bx-2

x-3(1)x+3x2

-5x+6=

A(x-1)+B(x-1)2

A(x

-1)

+B

”

x

+2(*)在(*)中分別令x

=1,2得：

A+B

=4B

=3

A

=1,

B

=3B(2)

x

+2

=A +(x

-1)2

x-1

(x-1)2A+2B+2C

=1

A

=1

A

=1,

B

=-1,C

=1A

B

C

A(x-1)2

+Bx(x-1)+Cx=

+

+x

x-1

(x-1)2

=

x(x-1)221(3)x(x-1)

A(x

-1)2

+Bx(x

-1)

+Cx”1(*)在(*)中分別令x

=1,0,2得：

C

=12A+B+C

=2

A

=1,

B

=-1,C

=12A+B-C

=0

A(x2

+1)

+x(Bx+C)

”

x

+1(*)在(*)中分別令x

=0,1,-1得：

A

=1A

Bx

+CA(x2+1)

+x(Bx

+C)x(x2

+1)(4)

x+1

x(x2

+1)

=

x

+

x2

+1

=化為基本分式及整式積分的代數和.5、有理函數R(x)的不定積分：u

+1如：

xudx=

1

xu

+1

+

C(u

?

-1)；adx

=

A(

x

-

a)

(x

-

a)aA d

(

x

-

a)；dxAx

+

Bd(x2

+

px

+

q)x2

+

px

+

q

x2

+

px

+

qdx

=

A1

+

B1

x2

+

px

+

q12p

2dxd(x2

+

px

+

q)+

Bx2

+

px

+

q(x

+

)+

k=

A1

；等.2x

-5x

+6例2、求不定積分：(x

-1)2x(x

-1)22x(x

+1)(1)

x

+3

dx

=

-5

+

6

dxx

-2

x

-3

2dx(2)

x

+2

dx

=

1

3

+

x

-1

(x

-1)

2(3)

1

dx

=

1

-1

1

+

+

dxx

-1

(x

-1)

x(4)

x

+1

dx

=

1+

-x

+1dxx

1+

x2

x-2

x-3=

(

-5

+

6

)dx=-5ln|

x-2|

+6ln|

x-3|

+Cx

-5x+6x+32(1)

dx例2、求不定積分：(x-2)56=ln

|

(x-3)

|

+Cx-2

x-3=-5d(x-2)

+6

d(x-3)

(x-1)x+2

dx2(2)3(x-1)2x-1+

]dx=

[

1=ln|

x-1|

-3(x-1)-1+Cx-1=d

(x-1)

+3

(x

-1)-2

d(x

-1)

x(x-1)1(3)

2

dx]dx1(x-1)2x

x-1=

[1

+

-1

+x-1=ln|

x

|

+(x-1)-1+Cx

x-1=dx

-d(x-1)

-

(x-1)-2

d(x-1)

x(x

+1)x

+1

dx(4)21+x21

-x+1x(

+

)dx=1

x

1

x

1+x2

1+x2-

+

)dx=

(2d

(1+x

)1+x2

dx

1+x2=dx

-

1x

2+

1+x2=ln

|x|

+arctanx

+C二、三角函數有理式的不定積分：A

=

R(sin

x,

cos

x)dx22dt1+t2設t

=tan

x

則：x

=2arctant

dx

=2t

1-t2且sin

x

=

,

cos

x

=1+t2

1+t2萬能代換法求A：1+sin

x

+cos

x例3、求不定積分：A

=

dx

=

1

dt

=ln|1+t

|

+C

=

ln

|1+

tan

x

|

+C1+t

22

2t

1-t1+t21+t2A

=1+t21+

+

2

dt2+2t

1

2

dt =22dt1+t2解：令t

=tan

x

則：x

=2arctant

dx

=2t

1-t2且sin

x

=

,

cos

x

=1+t2

1+t2詳見：變元積分法中根代換、三角代換、倒代換(1).注：若能用前邊所學方法解題則盡量不用本節(jié)所講方法.三、無理函數的不定積分：小結：1、部分分式法求有理函數的不定積分.2、萬能代換法求三角函數有理式的不定積分.3、變元代換法求無理函數的不定積分.4、靈活選用所學方法解題.作業(yè)4-4：218頁(6,8,20,21,24)(1)d

f

(x)dx

=

f

(x)dx(2)dF(x)

=F(x)+C.1、F'(x)=