數(shù)分1到不定積分

法，還有一些常用的公式。在本章里使用的積分公式除了161頁給出的10個常用公式外，還有6xx xx

x2

x2 arcsinxa2 x2 lnx2xx2x

2dx

a2lnxx

xax2x2 xa2xa2

a2x2dx

第一節(jié)定積分的 1(x5x3x)dx x6 x4 x2C x6 x4 x2 3* 4(5x)3dx(5x)3d(5x)1(5x)44 2 3 3xx3 )dx(x2x33x22x3)dx x23xx3

x36x23x3 1 1 x4(1x2)( 1x2)dx dx1x2dx x)dx11x3x1arctanx3 1x2dx(31x2)dx3x3arctanx

x xdx(x2x

2)dx

3x2x2x23

(2sinx4cosx)dx2cosx4sinx (3sec2x)dx (3 cos2

)dx3xtanx sin2x3cos2

x3)dx

cos2 dx(cos2

2)dxtanx2x2sin2xdx3cos2xdx3tanxxcos2

cos2sin

tan

dx

cosxdx

d(cosx)

cos2xsin2 cos dx

cos cos2cos2xsin2

coscosxsin

cosxsinxdx(cosxsinx)dxsinxcosx 1tanxC.1cos 1cos2xsin2 2cos2 2

52 (51)dx

dx 2ln ln

x 1

(2(3

)dx x ln ln3 xxexx

e

)dx(ex

)dxex

xx

(cosx )dxsinx2arctanx1arcsinx1 11

1

44x74

xxdxx2x4dxx4dx

223dx12dx

14 sinx)dx sinx)dx3arcsinx14 求一曲線yf(x),它在點(x,f(x))處的切線的斜率為2x,且過點(2,解：設f'(x2xf(x)f'(x)dxC2xdxCx2令f(2)5那么4C5,于是有C1,因此函數(shù)曲線yx21已知f(x)滿足給定的關(guān)系式,試求fxf'(x)1,(x1

f'(x)1,(xx解：可得f'(x)

x

解：可得f'(xx1dxlnxC.f(x)x

f(x)

xdx2f(x)f'(x)1,(x解：可知[f(xf'(x)]dx

f'(x)1,(f(x)f于是

f2

f'(xdx1dx,fC1xC2 2x因此有f(x2x

ln[f(x)]CxC 因此有f(x第二節(jié)換元積分與分部積分法用湊微分法求下列不定積分

1 d(5x6)1ln5x6C.5x6 5x6 (1 )dxlnxln2x1C.x(12x) 2x1xx1 x

(x1 x

1 [(x1)2 (x1)2] 2 [(x1)2(x1)2]3311311

)dx

131(131(x31(131( 1(1(x333213arcsinx arcsin(3x)33213d(3x)1 3d(3x)1 32

2 2132

3

e2dx2

2d

x)2

16216 xex2dx1ex2d(x2)1ex2d(x2)1ex2 1exdx1exd(e)ln(1e) d(ex exex2e2x12ex(ex1)2ex1C. d(ex) exexe2x1e2x1arctan(e)tanxdxsinxdx d(cosx)lncosx

cos tan5xxsecxdxcos2

cosdx

tanxd(tanx) 6

x

12sinxdx

dx 2sinxdxtanx

d(cosx)C1tanx

cos2 cos2

cos2 cos21

cos

cos2x d(tan d(tanAsin2xBcos2 Atan2x Atan2x BB d(Atan B B A(Atanx)2B

tan2xAtanx)C.cos5xdxcos4xd(sinx)(1sin2x)2d(sinx)(12sin2xsin4x)d(sinsinx2sin3x1sin5x xd(xd(x cos2

d(tan x x1

sin

12

22cos22

d(tanx x

x

cos dx2cos2xdxcos2xd(2x) d(sin2x)lnsin2xsinxcos

sin

sin

sin2x

sinxcosxdx 1sin4

sin1sin4

d(sinx)

1sin4

d1

x)

2

x) dx d(x2)4 24

24

d(x24)

ln(x24)2

4x4dx24x4d

)4

1(x22

d ) ) 5

dx

(4

)dx4x d(3x)C4x7ln3x21 31 33x3x

3x

sin2xcos3xdx12sin2xcos3xdx1(sin5xsinx)dx1cos5x1cosx (ln 1 dx(lnx)d(lnx)3lnx (24)sinxx2sinxd(x)cosx

dx(arcsin

2d(arcsinx)1arcsin3x31arctanxdxarctanxd(arctanx)1arctan2x11 x x 2d(x)2dx x 2d(x)2 x)

x(111

11

1d(ex)arcsin(ex)11sinxdx2sin2xcos2x

x

xd(

x2x2cos)2x 2(sinxcosx)d(x)2(sinxcosx) 用換元積x2x22x2x2x22x22x22x2

x2a2dxxx2x x2

)dx

)dx x2x2

)dxa2

dx

a2lnx

2x2x22x2x2可參考第四章第一節(jié)T9(7),此題所得可以當做公式使x2x2xx2x2x2a2dx lnx

x2sin 4sin2 4sin2 44sin24 d(2sint)2cosx(2cos44sin24令I(lǐng)sin2tdtJcos2tdt,IJ(sin2IJ(sin2tcos2t)dt dttCJI(cos2tsin2t)dtcos2tdt sin2tC于是因

It1sin2tC 44

dx2tsin2tC2

) 4 4

x24 4444

44dx dx4

dx 1(2

x42 4arcsinx( 4x2 xC2arcsinxx42 x1 x5x dx dx dx1 d(x1)21(1x)2 21(x1)2 21(x5x d(x1)21 d[2(x1221(x1 1[2(x1 221(x1)21 (x12 5x 1arcsin2x5x2

2xx2dxx

d(x199(x1 22x22x

12x(2x(2x

9arcsin2x1

x2a2x2dx

x2 dx

a2

a2

a2

3(x2a2 (x2a2 (x2a2 (x2a2

12 d(x2a2a2

1dx2a (x2a2 dx

1 1a2 (x2a2

312

(x2a2

dx1

a2

a2 (x2a2 (x2a2 (x2a2 1(x2a2x x2x x2x x2

x2(x21)dx(x x1)dxxdxxx2( x211lnx x21) e etd(t21)2tetdt2td(et)2tet2etdtt2tet2et 2x1ex12ex1

1dxtx1t 1)(12)2tdt(2t4t)dt(2t44xx t1dxx

t t txtxt24t4lnt1 x1 4 x1)

t6x13x13

t1t2dt6t

1

2dt(6t66t46t26

1

26t76t52t36t6arctant t6x 7x7/6

x5/62x1/26x1/66arctan(x1/6)

t

dt2

dt(2t

2t2

2t3t22t2ln1t

1

1

1 2t 23

x3/

x 2 x)x dxx11

dx2t

(1t2t

d(1t

)(1t2

t

15t2t31t515

2(1x2)3

1(1

)212 12x2 (x1)22(x1)(12).(x1)3dx

d(x (x d(x1) d(xxlnx1

(x x 2(x

(x用分部積分法求下列不定積x2cos

x2dsinx2sinx

sinxdx2 4

ln xsinx2xsinxdx 4lnx4xd(lnx)x2sinx2xdcosx

x2sinx2xcosx2cosxdx 4 x

lnx x3dx1x2sinx2xcosx2sinx lnx x4 lnxdxxlnxxd(lnx)C1xlnxdxC1xlnxxarctanxarctanxxd(arctanx)x

arcsinx12arcsinxd(1x2arcsinx(1x) 1xd(arcsinx) xarctan

dx1 1 1

2arcsinx(1x)

1xdxxarctanx21x2dx21

2arcsinx(1x)

dxxarctanx

ln(1x2)2

12arcsinx(1x)1

xarctanxdx2arctanxdx2arctanx2xdarctanx 2arctanx21x2dxC12arctanx2(11x2)dxx2arctan

ln

dx1lnxd

1)

lnx

d(lnx)C

lnx

dx 2lnx

1

2 1 1cos(lnx)dxxcos(lnx)xd[cos(lnx)]C1xcos(lnx)sin[lnx]dxxcos(lnx)xsin(lnx)xd[sin(lnx)]C2x[cos(lnx)sin(lnx)]cos(lnx)dxcos(lnx)dxx[cos(lnx)sin(lnx)]2sec5xdxsec3xd(tanx)sec3xtanxtanxd(sec3x)sin

sin2sec3xtanx3tanx dxC

xtan

cos4

cos5

1cos2xdx

sec

3xdxC因此

cos5

xtan

3

3 sec5xdx

sec3xtanx3sec34

sec3xdxsecxd(tanx)secxtanxtanxd(secx) sin sin21secxtanxtanxcos2xdxC3secxtanxcos3xdx1 1cos2

3xdx dxsecxtanx cos3 dxC3secxtanx

cos secxtanxsec3xdxlnsecxtanxC3于sec3xdxsecxtanxlnsecxtanxC 綜上可 5xdxsecxtanx3(secxtanxlnsecxtanx 3

1 xln(1x)dxxln(1x)dxxln(1x)dx2ln(1x)d(x)2ln(1x)d(x1{x2ln(1x)2

x2d[ln(1x)]}1{x2ln(1x)2

x2d[ln(1x)]}1 1 1 2x ) )dxC1 x2 ) 2dx1 1 1 1 1 11 )

1)dxCx1

1x)

x

1 1 1 1x1x2ln(1x)1ln(1x) 1 1xsin2解2那么IJxdxx22

Ixsin2 Jxcos2JIx(cos2xsin2x)dxxcos2xdx1xd(sin xsin2x1sin2xdxCxsin2x2cos2x C于是

I xsin2xdxx2xsin2xcos2x

xsin

cos 2xcos2xdx,依上題可知xcos2xdx2

xsin

cos [ln(lnx)1]dxln(lnx)dxdxxln(lnx)xd[ln(lnx)]dxln ln lnxln(lnx)x1dxdxCxln(lnx)1dxdxlnxxln(lnx)

ln ln ln (x (x1)2dx (x dxx1dx(x1)2dxx1(x1)2ex ed(x) 2dxx (xex 1x (x1)2d(x1

)(x1)2dxCx1 arcsinxt t2d(sint)t2sintsintd(t2)Ct2sint2tsintdt t2sint2td(cost)Ct2sint2tcost2costdt1arcsin1t2sint2tcost2sint (arcsinx)2x arcsinx2xdx dxxd(cotx)xcotxcotxdxCxcotxdxsin2 sin xcotx d(sinx)dxCxcotxlnsinxsin 1ln(x1x2)dxxln(1

)xd[ln(x

)]1x1

)

x

1111

dx11xln(x ) dx11111xln(xxln(x

1x2)11

11 2 xln2xdx

xd(x2)

x2lnx

x2d(ln2x)32

2 x2lnx

2x2lnxdxC1 x2lnx lnxd(x2) 2 8 x2lnx x2lnx8x2d(lnx)C22x2lnx x2lnx8x2dx x2lnx x2lnx x2 求下列不定積分的遞推公式I

xnekxdx

xnd(ekx)1xnekx

ekxd(xn)1xnekx

ekxxn1dx1xnekxnI k

k

k

kIntannxdx

sin2xxcos2

dx

(1cos2 cos2

tann2cos2

dxtann2

tann2xdxtann1xn

n2I(arcsinx)ndxx(arcsinx)nxd(arcsinx)nx(arcsinx)n (arcsin 1 11x(arcsinx)nn(arcsinx)n1d(1x21x(arcsinx)n1x(arcsinx)n11x(arcsinx)n1

(arcsinx)n1n1x2d[(arcsinx)n1](arcsinx)n1n(n1)(arcsinx)n2dx(arcsinx)n1n(n1)In2.求下列有理函數(shù)的積分x5x4 x3 dx(xx

)dxx x1

8ln

4lnx

3lnx1

(1x1

(x1)(x2

x2 )dx 2x x x x1lnx211arctanx1lnx14

11112(3).1x4dx2

(x21x2)dx4

1( 1(

x2)dx1lnx1 x dx1

3x x2x

3x2x 1lnx1

x dx

1) 3

(x

1)2(3 d(x1 lnx1 d(x1) 6

x1)23

1 3

(x ) 1lnx1

1lnx2x112arctan[2(x1)] 1313 131331lnx11lnx2x1 arctan(2x )3 x x x dx d(x

7)

1lnx27x12

x7

7x

(x7)2 xxx

x7212121lnx27x1231212

C2lnx4lnx32(x dx (

4)dx3lnx15lnx12lnx2 (x21)(x

6x x x 2424242224242422x

1 1

2)dx

(x

(x2222)42222)

)4)x4

xx222xx22x2xx222xx22x2

(x 2)2 (x2)2 2 ln arctan(2x1) arctan(2x1) (x2)2 2ln

x2 2xx22xx2 2xx22x2

arctan(2x1)2422x

dx 2(x1)5dxln(x22x1)

5x22x

(x1)

x22 d(x2)1lnx222

2xxx22xx

(x2)2

x2( (

dx dx)1 82x 6(x2)(x 6

x x 求下列三角函數(shù)有理式積分 d(2

d(45cosx2 x 4 5 9 2xd(

d(tan2

2 9

9

x21

cos24sin

5sin2x5cos2x8sinxcos d(tan 1

5tan2x58tand(tan 1

d(tan5tan2x58tan

tan2x8tanx

5(tanx4 )15 5(tanx

arctan C arctan(tan 53

sin2

d(tanx) 13tan2x

d(tanx) tanx)32 2tan32 2366 66 1

tanx)

sec

2dx

cos dx

dx

cos

2 sec

(cosx

cos

(cosx1)(1 cosx

dx dx

d()

x

d(xcosx (cosx

2

(2 d(tan2 tan tan22cos2xC1tan22cos2x2tan2d(cos2x) tan 2

3

2dxC

tan

2 2 2 2 22 2 cosx 1tanx cosx 取I cos cosxsin J sin dx

cosxsinIJdxx cosxsin d(cosxsin IJcosxsinxdx

lncosxsinxcosxsinxlncosxsinIxlncosxsinlncos cosxlncossin

x 2sin J

dx cosxsin 1(sinx2cosx)dx1d(cosx)2dx1d(sin(2cosx)sin

132cos

312cos sin sinln2cosx3

1cos2

lnsinx 13

122cossin

1cos1cos sinxcosxdx1(12sinxcosxdx sinx1cos1cos xx 2 1((sinxcosx)dx sinxcos

x1tan

22tanx22sinx cosx ln C.sin3

sin2

tanx21cos2 1cos2xdx1cos2xd(cosx)1cos2xd(cosx)(11cos2x)d(coscosx2arctan(cosx)參見T4(3) tan3 tan2xtanxdxC1 tan2xlncosx dx cos dx dsin sin sinxsinx sin2xcos sin2xcos2 sin2x(1sin2 t2(1t2)(t2sinxsinx 11

sinttt

1 sin 取

cos sinx2cos

I cos sinx2cos J sin dx2IJdxx

sinx2cos cosx2sin I2J

sinx2cos

dx

sinx2cos

lnsinx2cosxsin2

I cosx dx2x1lnsinx2cosxC.sinx2cosx sin2 1cos2xdx2sin2xdx(12sin2x)dxx22sin2xx d(tantan2x

x 2arctan(2tanx)2 x1/6 6tx(12x3x1 t6(12t3t2)6t(1t)(2t2t1)2(tt12t2x(12x3x16lnt

3ln1t

6(t)6(t) 4

(t

1)2 34346lnt3ln1t9lnt2t1

arctan[47(t1)]

3ln1x1/69lnx1/3

37arctan[47(x1/6lnx

)]x1 xx1

xxdx (xxx2

xdx

xx1x12x2x2x2(x

x21)dxx2 x211lnx

2d(211

t2t2t2

xx2xx22x

x212 2

221

2d(t12 (t1)2 1x1x11

lnt1 2t22t2t2x 2

1xx22x2dx[(x1)(x1)21 (x1)21 (x1)21d(x1)22

(x1)2x22x 1(x22x2)3x x22x21x22x 1x1xx1 1x1xx1

1111112

d 11

d(1 1

ln111

x2(x1)2 1x2(x1)2 1x22

11x

dx

1x11x

dx

111x

dx

(x1)1xx2dx 25(x1 x 2(x

2(x1 25 x

21ln1x 2(x1

32

25 211 21ln1xx2 x2x)3x2x)3

x1dx

(x12dx x2x13lnx

s41 x2t dts41x41x241x241t41tx41x241x241t41tt241t11y41s1d(y4 11 4(y41)y(y41)141141y4141141

141t141t141t141t 14141141

C x2

1arctan41arctan41

d(114

x

4x3(14x3(1xx

(1t4t22t22tt22t 22 2t)22 2t) 281t

14x14x14 14 )3/ 22241241 24 x11

21x)224 x4

1 4 x2 2

](xa)(x(a(xa)(x(ab)2(xab42(ab)2ab(xab 2(xabarcsin Carcsin2xabC.ab ab1 d(x2

d(x

11x2x2xexsinxdxxexd(cosx)xexcosxcosxd(xex)xexcosxcosx(exxex由于xexsinxdxxsinxdexxsinxexexd(xsinx)xsinxexex(sinxxcos于是可以求

xexsinxdxxexcosxdxxexcosxcos ex(sinxcoscosxedx

ex(sinxcos于是

cosxexdx xexsinxdxxex(sinxcosx)exsinx

xex(sinxcos excosxexcosxdx xexcos

sin(b sin(xbxsin(xa)sin(x sin(ba)sin(xa)sin(x sin(ba)sin(xa)sin(x sin(xb)cos(xa)sin(xa)cos(xb)sin(ba)sin(xa)sin(x cos(x cos(xsin(ba)sin(x sin(ba)sin(x sin(b

lnsin(xa)

lnsin(xb) sin(ba)

sin(xsin(xa)sin(x tanxtan(xa)dx sinxsin(xa)dx[cosxcos(xa)sinxsin(xa) cosxcos(x cosxcos(xx

coscosxcos(x

dx

x

coscos2xcosasinxcosxsin

xcosad(tanx)

xcosalnsinatanxcosacosatanxsin

sin

x3arccos

dx2

x2arccos

d(x2)

(1x2)arccos1 1111

arccos11

d(x2111 arccosxd(1x2)1arccosxd11 1 arccosxd(1x2)2arccosxd(1x2)2331arccosx(13

)2

(1x2)2d(arccos arccosx(1x2)1x2d(arccosx)