不定積分練習(xí)題答案

名主要內(nèi)f(x，xI，若存在函數(shù)F(x，使得對任意xI均有F(xfdF(x)f(x)dx，則稱F(x為f(x)的一個原函數(shù)f(x)的全部原函數(shù)稱為f(x)I上的不定積分，記f(x)dxF(x)注：（1）若f(x)連續(xù)，則必可積；（2）F(xG(x)均為f(x)的原函數(shù)，則F(x)G(x)C。故不定積分的表達式不唯一d 性質(zhì) f(x)dxf(x)或 f(x)dxf(x)dxdx 性質(zhì)2F(x)dxF(xC或dF(x)F(xC性質(zhì)3[f(xg(x)]dxf(x)dxg(x)dx，為非（法設(shè)f(u)的原函數(shù)為F(u)，u(x)可導(dǎo)，則有換 f((x))(x)dxf((x))d(x)F((x))x(t單調(diào)f[(t)](t有原F(t則f(x)dxf((t))(t)dtF(tCF(1(xu(x)v(x)dxu(x)dv(x)u(x)v(x)有理函數(shù)積分若有理函數(shù)為假分式，則先將其變?yōu)槎囗検胶驼娣质降暮?；對真分式的xxxx

2，由積分表中 x解x

1

2dx3

x23★(2)3

11x解3xx

1)dx(x3xdxlxdxl★(4)

x(x解

x(x3)dxx2dx3x23x43x2★★(5)

x2

3x43x2x2

3x2

x2

解

3x43x2x2

dx3x2dx2x★★(6)1x22思路:注意到2

x211

1

1

1 解1x2dxdx1x2dxxarctanx 1 dx解 13-4dx

1dx3x3dx

2 x3

2 （-1x2ln|x|3x24x3 111

111解：

dx

dx3arctanx2arcsinx★★(9)

1xxx

1 xx11 xxxxx

？看7

x248

x8，直接積分解

xxxdx x8dx x1

★★(10)x2(1x21111 (解： (x(1xe2x

1

)dx

dx

1

dx1arctanxx★(11)ex1e2x (ex1)(ex 解ex1dx★★(12)3xex

ex

dx

1)dx

xx

（3x★★(13)cot2思路:應(yīng)用三角恒等式cot2xcsc2x1解cot2xdx(csc2x1)dxcotxx23x5★★(14)

23x5

2

2x 3x23x5 2 (解

2

x2x ln2ln★★(15)cos2 解：cos2xd1cosxdx1x1sinx 1 1★★(16)1cos解： dx

dx

sec2xdx1tanx1cos 2cos2 2 (17) cosxsin思路:不難，關(guān)鍵知道cos2xcos2xsin2xcosxsinx)(cosxsinx 解： cos2x dx(cosxsinx)dxsinxcosxC.cosx ★(18)cos2xsin2

cos2xsin2 解cos2xsin2xdxcos2xsin2xdxsin2xdxcos211csc2xdxsec2xdxcotx11111111111

1x111

11解

1x)dx2 dx2arcsinx111111cos2★★(20)1cos1cos2

1cos2

1cos

2cos2

x ，則積分易得21cos2 tanx解1cos2xdx2

xdx2dx

2解(1)dx1d(7x3);(2)xdx1d(1x23)x3dx1d(3x4 (4)e2xdx1d(e2x);(5)dx1d(5ln|x|);(6)dx1d(35ln|x t(7)1dt2d(t 1d(tan 1d(arctant

cos2

1 ★（1）e3t 解：e3tdt1e3td(3t1e3t 3★(2)(35x)3 解：(35x)3dx1(35x)3d(35x)1(35x)4 1★(3)31解： dx1 d(32x)1ln|32x|3 23 35★(4)35 解

dx35

d(53x)35

(53x)3d(53x)

(53x)3x★(5)(sinaxeb 解(sinaxeb)dx

sinaxd(ax)bebd()

cosaxbeb

tt思路:如果你能看到dt

t)

12dt12

t易解解：t

tdt2

td

t)2 ★(7)tan10xsec2 解：tan10xsec2xdx tan10xd(tanx)1tan11x xlnxlnln解： d(ln|x|)

d(ln|lnx|)ln|lnlnx|xlnxlnln lnxlnln1111

lnln11

1是什么，是什么呢？就是11x21x21

tan1x2

ln|

|111★★(10)sinxcos sin2x2sinxcosx 2dxcsc2xd2xln|csc2xcot2x|Csinxcos sin方法二：將被積函數(shù)湊出tanx的函數(shù)和tanx的導(dǎo) sinxcos

dx sinxcos

tan

sec2xdx

tan

dtanxln|tanx|方法三：三 sin2xcos2x1，然后湊微分

sin2xcos2

sin

dcos dsinsinxcos

sinxcos

dxcosxdxsinxdx

cos

sinln|cosx|ln|sinx|Cln|tanx|★★(11)exe ex 思路:湊微分exe

e2x

1

1

)2 ex 解exe

e2x11(ex

arctane★(12)xcos(x2 解：xcos(x2dx1cosx2dx21sinx2 22

1d(23x2222262

解

1

d(23x2)

(23x2

2d(23x2)

23x22222★★(14)cos2(t) 解cos2tsin(t)dt1cos2tsin(t)dt1cos2t)d 1cos3(t)★★(15)1x4

解：1x4dx41x4dx41x4

41x4d(1

) ln|14

|

sinx cos 解：sinxdx

dcosx

cos3

cos3

cos2★★(17)

211解

20dx★★(18)

1 999解：1 dx991 d2x1 32x1 d22x

2x2

1arcsin(2x)

9解 2x2

(2x1)(1 2 2x21 d(2 2x2★(20)(4(4(4

145x4）dx1（

）d(4（5(425（5(4254(4

d(45x)

d(45x)

1ln|45x|

254

25(4

254★(21)(x

(x11)2

(x

(x 解(x1)100

(x

((x1)1002(x1)100(x1)100

(x

(x

(x

)d(x1 97(x1)97

1 49(x1)98

1 99(x1)99

★★(22)x81解 xdx

1

)xdx1

x8

(x41)(x4

2x4

x4 4x4

x41

44[2(x214

)

8[

1d

1d

x21 14(x2)21★(23)cos3

ln|

x2

| 4

思路:湊微分。cosxdxdsinx解cos3xdxcos2xcosxdxcos2xdsinx(1sin2x)dsinsinx1sin3x3★★(24)cos2(t 解：cos2(t)dt1cos2(t)dt 1dt1cos2(t)d2( 1t1sin2(t) ★★★(25)sin2xcos解sin2xcos3xdx1(sin5xsinx)dx

sin5xd5x1sin 1cos5x1cosx ★★★(26)sin5xsin 解：sin5xsin7xdx 1(cos2xcos12x)dx1cos2xd2x1 1sin2x1sin12x ★★★(27)tan3xsec思路:湊微分tanxsecxdxdsecx解tan3xsecxdxtan2xtanxsecxdxtan2xdsecx(sec2x1)dsec3sec2xdsecxdsecx1sec3xsecx3★★(28)

10arccos111

dxd(arccosx解

10arccos 1

darccosx

10arccos

(arcsinx)21(arcsinx)2111

dxd(arcsinx解 darcsinx (arcsinx)21 (arcsinx)21

xdx

2 12 1(xx

xd

xxxx解： dx2arctan dxxx

2

xd x 1(x x)2lntan★★★★(31)cosxsinx思路:被積函數(shù)中間變量為tanx，故須在微分中湊出tanx，即被積函數(shù)中湊出sec2x

dx

lntan

dxlntanxsec2xdxlntanxdtancosxsin

cos2xtan

tan

tanlntanxd(lntanx) 解 lntan

dx

lntan

dx

d((lntanx))lntanxdtanx

cosxsin1(lntanx)22

cos2xtan

tan 1ln★★★★(32)(xlnx)21思路d(xlnx1ln1 解：1lnxdx (xln

(xln

d(xlnx)

Cxlnx★★★★(33)1思路:將被積函數(shù)的分子分母同時除以ex，則湊微分易得1

eexe

dx

d

)

d

1)ln|

1| 1

1ex1

edx1dx1exe

x 1d(1ex 1 xln|1ex|Cxln(ex|ex1|)x(lnexln|ex1|)Cln|ex1|思路:將被積函數(shù)的分子分母同時乘以ex，裂

1

1

ex(1ex

ex(1ex

1ex

ln

1

d(1exln|1ex|Cln|ex1| x(x6 x4xdx11 x(x6 4x(x6 4 x(x6 4 x64

d(x6 4ln|x|24

x6

ln|x| ln|

4|1令x ，則dx1dt1 t2

d(4t6)

d(4t6x(x6

t6

(t2

241

24

11ln(14t6)C1ln(14) x8(1x2x8

1x8

(1x2)(1x2)(1x4

2

x(1

dx)

x(1x

dx

1 1x2x4 x

dx 1(1x)(11 (1 1111

1

)dx

1x2

1

11

方法二：思路:1令x ，則dx1dt1 t28

x8(1x2

1t

(t2dt)t2

dt(tt1

1t

(t6t4t21)dt )dt

(t6t4t21)dt1(1

1 t21

2t

t1t71t51t3t1ln|t1|C1

1

1111ln|1x| t 7

5

3

1

sin2xcos2x sec2xtan2x1111思路:xsint,

，先進行三角換元，分項后，再用三角函數(shù)的升降 2解xsint,

2 costdtdt t tsec2tdt1 111111

1cos

2cos2 2t Carcsinx C.（或arcsinx C 11 （

tant

sin

1cos1 ，又sintx時，cost1★★★(2)

x2xx2x

sin思路:x3sectt

2

解x3sectt0dx3secttantdt2xx dx 3secttantdt3tantdt3(sectx2x23tant3tC

|x

（x3secx時cosx

3,sinx

,tanx x2x2x2x2(x2★★★(x2思路:xtant,解xtant,

2dxsec2tdt2

sec2tdt

dt (x2 (x2

costdtsintC 1(x21(x2a2思路:xatant,

，三2解xatant,

dxasec2tdt2

asec2tdt

(x2a2 (x2a2

a2sec a2

costdt

sinta2a2x2xxx2xx4思路:先令ux2，進行第一次換元；然后令utant,

x2x4x2xx4x2xx42

，令ux2得x2 u xx41dx2uu21du，令utant,

，du2

tdtx2xx4 dx1u1du1tant1sec2tdt1x2xx4u2 tantu2

1(csctsect)dt1lnsecttant1lncsctcott u2u2u1ux4u2u2u1ux4x41 u

C

x

★★★(6)

54xx2解：54xx29x2)2x23sint,

，則dx3costdt2 54xx2dx9cos2tdt91cos2tdt9(t1sin2t) x 54xx 54x

1、求下列不定積分★（1）arcsin思路:被積函數(shù)的形式看作x0arcsinx，按照“反、對、冪、三、指”順序，冪函數(shù)x0優(yōu)先納入到微分dx。112解：arcsinxdxxarcsinxx dxxarcsinx1 d1121xarcsin1★★（2）ln(1x2

解ln(1

2)dxxln(1

)x2xdxxln(1 1

)1x2xln(1

)

2(x21)1

dxxln(1

)2dx 1 xln(1x2)2x2arctanx★（3）arctan

d(1x2解arctanxdxxarctanxx1x2xarctanx2xarctanx1ln(1x2)2

1★★(4)

sin2 解 e2xsinxdxsinxd(1e2x)1e2xsinx1e2x1cos 1e2xsinx1cosxd(1e2x 1e2xsinx1(1e2xcosx1e2xsin 1e2xsinx1e2xcosx1e2xsin e2xsinxdx2e2x(4sinxcosx) ★★(5)x2arctan 1 1 解

arctanxdxarctanxd(3)3

arctanx3

1x21

x3x

1 3xarctanx3

1

xarctanx3(x1x21x3arctanx

xdx dx1x3arctanx1x2

d(1x2 3

31

611x3arctanx1x21ln(1x2) x★(6)xcosx 解：xcosxdx2xdsinx2xsinx2sinxdx2xsinx4 22xsinx4cosx ★★(7)xtan2解xtan2xdxx(sec2x1)dx(xsec2xx)dxxsec2xdx xd(tanx)xdxxtanxtanxdx1x2xtanxlncosx1x2 ★★(8)ln2 解：ln2xdxxln2xx2lnx1dxxln2x2lnxdxxln2x2xlnx2x1 xln2x2xlnx2dxxln2x2xlnx2x★★(9)xln(x 1 1解xln(x1)dxln(x1)d22xln(x12x11

ln(x1)

x211

1

ln(x1)

1(x

1 x x1x2ln(x1)1x21x1ln(x1) ln2★★(10)x2ln2ln221)1212lnx1dx12lnxxxxx

dx

lnx

lnx 1ln2x2lnxd(1)1ln2x2lnx 1dx1ln2x2lnx2 1(ln2xlnx2)C★★(11)coslnx解：coslnxdxxcoslnxxsinlnx1dxxcoslnxsinlnxxxcoslnxxsinlnxxcoslnx1dxxcoslnxxsinlnxcoslnxcoslnxdxx(coslnxsinlnx)2★★(12)

lnxdx思路：詳見第(10)★★(13)xnln

(n解

xlnxdxlnxdn1x

n

lnx

n

xn1

xn1lnx

xndx

xn1lnx

n

n

n

(n1) ★★(14)解x2exdxx2exex2xdxx2ex2xex2x2ex2xex2exCex(x22x2)★★(15)x3(lnx)2解x3(lnx)2dx(lnx)2d(1x41x4lnx)21x42lnx1 1x4(lnx)2 x3lnxdx1x4(lnx)21ln 1x4(lnx)21x4lnx1x41dx1x4(lnx)21x4lnx1 1x4(lnx)21x4lnx1x4C1x4(2ln2xlnx1) lnln★★(16) 思路達式lnlnxdx寫成lnlnxd(lnx)，將lnxx解lnlnxdxlnlnxd(lnx)lnxlnlnxlnx11dxlnxlnlnx1xlnxlnlnxlnxClnx(lnlnx1)

lnx ★★★

xsinxcos 解：xsinxcosxdx 1xsin2xdx xd(1cos2x)1xcos2x 1xcos2x1cos2xd2x1xcos2x1sin2x 2★★(18)

2思路：先將cos2x降冪得1cosx，然后分項積分；第二個積分嚴(yán)格按照“反、對、冪、三、指” 解：x2cos2xdx(1x21x2cosx)dx x2dx 1x3 x2dsinx1x31x2sinx12xsin 1x31x2sinx xdcosx1x31x2sinxxcosxcos 1x31x2sinxxcosxsinx ★★(19)(x21)sin 解(x21sin2xdxx2sin2xdxsin2xdxx2d(1cos2x1cos 1x2cos2x12xcos2xdx1cos2x1x2cos2x xdsin 1cos2x1x2cos2x1xsin2x1sin2xdx1cos 1x2cos2x1xsin2x1cos2x1cos2x 1x2cos2x1xsin2x3cos2xC1(xsin2x3)cos2xxsin2x ★★★(20)e3x3解：令t ，則xt3,dx3t3e3xdxet3t2dt3ett2dt3t2det3t2et33t2et32tdet3t2et6ett6etdt3t2et6ett6et33x2e3x6e3x3x6e3xC3e3x(3x223x2)★★★(21)(arcsin1思路nx)2xxnx)2x2n11x(arcsinx)2arcsinxd(1x2)x(arcsinx)22arcsinxd(1x11x(arcsinx1

arcsinx

111111★★★(22)exsin2

arcsinx2dxx(arcsinx)2

arcsinx2x11exsin2xdxsin2xdexexsin2xex2sinxcosexsin2xexsinexsin2xdxsin2xdexexsin2xex2cos2xdxexsin2x2cosexsin2x2excos2x4exsinexsin2xdxex(sin2x2cos2x)5exsin2xdxex(5sin2xsin2x2cos2x)5 exsin2xdxex1cos2xdx1exdx1excos2xdx1ex1 excos2xdxcos2xdexexcos2xex2sin2xdxexcos2x2sinexcos2x2exsin2x4excosexcos2xdxex(cos2x2sin2x)5exsin2xdxex1exsin2x1excos2x x★★★(23)ln(1xx解ln(1x)dxln(1x)dx

x

xln(1x)2x1x令t x，則dx2 t21xdx41t2dt4dt41t2dt4t4arctantxx 4 xxxxx所以原積分xxx

dx

xln(1x)

4

C★★★(24)

ln(1ex解

dxln(1

)

)

1ex

xln(1

e)1exdxe

xln(1

) d(1ex)1e exln(1ex)ln(1ex) 注：該題 1

1★★★(25)xln1dd x) x 解：xln1xdx

ln1 1 1 1

x21x1x1x 1

1 1 2

1 (11 1 1 1 2xln1x1x2dx2xln1xdx1x21x2ln