- 1 -

- 1 -

- 1 -

第三章 不定积分 一. 求下列不定积分:

1. ∫ −+

−dx

xx

x 11ln

11

2

解. =−+

−∫ dxxx

x 11ln

11

2 cxx

xxd

xx

+

−+

=−+

−+

∫2

11ln

41

11ln

11ln

21

2. ∫ −+

+dx

xx

x 11arctan

11

2

解. ∫ ∫∫ −+

−+

=−+

=−+

+ xxd

xxxd

xxdx

xx

x 11arctan

11arctanarctan

11arctan

11arctan

11

2

= cxx

+

−+ 2

11arctan

21

(直接计算可知xxdxd

−+

=11arctanarctan ).

3. ∫ ++

⋅+

++ dxxx

xxx

cos1sin1

)cos1(1sincos

2

解. cxx

xxd

xxdx

xx

xxx

+

++

=++

++

=++

⋅+

++∫∫

2

2 cos1sin1

21

cos1sin1

cos1sin1

cos1sin1

)cos1(1sincos

4. ∫ + )1( 8xxdx

解. 方法一: 令t

x 1= , ct

tdttdt

tt

txxdx

++−=+

−=

+

−=

+ ∫ ∫∫ )1ln(81

1111

1

)1(8

8

7

8

2

8

= cx

+

+− 8

11ln81

方法二: ∫ ∫∫ +−−=

+=

+dx

xxx

xxdxx

xxdx )

111(

)1()1( 887

88

7

8

= cxxxxd

xdx

++−=++

−∫ ∫ )1ln(81||ln

1)1(

81 8

8

8

= cx

+

+− 8

11ln81

5. ∫ +++ dx

xxxcossin1

sin1

解. ∫ ∫∫+

=+

+

=++

+ dxx

xx

dxxxx

xx

dxxx

x

2cos2

2cos

2sin

2cos2

2cos

2sin2

2cos

2sin

cossin1sin1

2

2

- 2 -

= ∫ ∫ +−=− cxxx

xddx |

2cos|ln

21

2cos

2cos

21

二. 求下列不定积分:

1. ∫+++ 22)1( 22 xxx

dx

解. ∫∫+++

+=

+++ 1)1()1()1(

22)1( 2222 xxxd

xxxdx

tx tan1 =+令 ∫ ttt

dt

sectancos

2

2

= ∫ ++

++−=+−= c

xxxc

tttdt

122

sin1

sincos 2

2

2. ∫+ 24 1 xx

dx

解. 令 x = tan t,

∫ ∫ ∫ ∫∫ ++−=−===+

cttt

tdttddt

tt

ttt

dt

xxdx

sin1

sin31

sinsin

sinsin

sincos

sectancos

1 3244

3

4

2

24

= cx

xx

x+

++

+−

23

2 1131

3. ∫++ 22 1)12( xx

dx

解. 令 tx tan=

∫ ∫∫∫ +=

+=

+=

++ ttddt

tttdt

ttt

xxdx

2222

2

22 sin1sin

cossin2cos

sec)1tan2(sec

1)12(

= cx

xct ++

=+21

arctansinarctan

4. ∫− 22

2

xadxx

(a > 0)

解. 令 tax sin=

∫ ∫∫ +−=−

=⋅

=−

ctatadttata

tdtataxa

dxx 2sin41

21

22cos1

coscossin 222

22

22

2

- 3 -

= cxaax

axa

+

−− 22

2

2

arcsin2

5. ∫ − dxx 32 )1(

解. 令 tx sin=

∫∫ ∫∫++

=+

==− dtttdtttdtdxx4

2cos2cos214

)2cos1(cos)1(22

432

= ∫ +++=+++ ctttdtttt 4sin3212sin

41

83)4cos1(

812sin

41

41

= cttx +++ )2cos411(2sin

41arcsin

83

= ctttx +−+

+ )4

sin214(cossin241arcsin

83 2

= cxxxx +−−+ )25(181arcsin

83 22

6. ∫− dx

xx

4

2 1

解. 令t

x 1=

∫ ∫∫ −−=

−

−

=− dtttdt

tt

tt

dxx

x 22

4

2

2

4

2

111

11

ut sin=令 ∫− uduu 2cossin

= cx

xcu +

−=+ 3

323

3)1(

cos31

7. ∫−

+ dxxx

x1

122

解. 令t

x 1=

∫ ∫∫−

+−=

−

−

+

=−

+ dtt

tdtt

tt

t

tt

dxxx

x22

2

2

2

22 111

11

1

11

ut sin=令 ∫+

− uduu

u coscos

1sin= c

xxxcuu +−

−=+−

1arcsin1cos2

- 4 -

三. 求下列不定积分:

1. ∫ +−+ dxee

eexx

xx

124

3

解. ∫ ∫∫ +−=+−

−=

+−+

=+−

+ −−

−

−

−

ceeee

eeddxee

eedxee

ee xxxx

xx

xx

xx

xx

xx

)arctan(1)(

)(11 22224

3

2. ∫ + )41(2 xxdx

解. 令 xt 2= , 2lnt

dtdx =

ctt

dttttt

dtdxxx +−−=

+−=

+=

+ ∫∫∫ 2lnarctan

2ln1

111

2ln1

2ln)1()41(2 2222

= cxx ++− − )2arctan2(2ln

1

四. 求下列不定积分:

1. ∫ −dx

xx

100

5

)2(

解. ∫∫∫ −− −+−

−=−−=−

dxxxxxxdxdx

xx 994

99

5995

100

5

)2(995

)2(99)2(

991

)2(

= ∫ −−⋅⋅

+−×

−−

− dxxxx

xxx 983

98

4

99

5

)2(989945

)2(98995

)2(99

= 96

2

97

3

98

4

99

5

)2(96979899345

)2(97989945

)2(98995

)2(99 −⋅⋅⋅⋅⋅

−−⋅⋅

⋅−

−⋅−

−−

xx

xx

xx

xx

cxx

x+

−⋅⋅⋅⋅⋅⋅⋅⋅

−−⋅⋅⋅⋅

⋅⋅⋅− 9495 )2(9596979899

2345)2(9596979899

2345

2. ∫+ 41 xx

dx

解. ∫ ∫∫ ∫+

−=+

−=+

−=

+ 22

2

4

4

4

2

4 )(121

111

1

/11 t

dtt

tdt

tt

t

dtttx

xxdx

令

- 5 -

cx

xcuuduuuut +

+−=++−=−= ∫ 2

422 1ln

21|sectan|ln

21

secsec

21tan令

五. 求下列不定积分:

1. ∫ xdxx 2cos

解. ∫ ∫∫ +=+= xxdxdxxxxdxx 2sin41

41)2cos1(

21cos 22

∫−+= xdxxxx 2sin412sin

41

41 2

cxxxx +++= 2cos812sin

41

41 2

2. ∫ xdx3sec

解. ∫ ∫∫ −== xdxxxxxxxdxdx tansectantansectansecsec3

= ∫∫ −++=−− xdxxxxxxdxxxx 32 sec|tansec|lntansecsec)1(sectansec

cxxxxxdx +++=∫ |tansec|ln21tansec

21sec3

3. ∫ dxxx2

3)(ln

解. ∫ ∫∫ +−=−= dxx

xxxx

dxdxxx

2

233

2

3 )(ln3)(ln11)(ln)(ln

∫+−−= dxx

xxx

xx

2

23 ln6)(ln3)(ln∫+−−−= dx

xxx

xx

xx

2

23 6ln6)(ln3)(ln

cxx

xxx

xx

+−−−−=6ln6)(ln3)(ln 23

4. ∫ dxx)cos(ln

解. ∫∫∫ −+=+= dxxxxxdxxxxdxx )cos(ln)]sin(ln)[cos(ln)sin(ln)cos(ln)cos(ln

= cxxxdxx ++=∫ )]sin(ln)[cos(ln2

)cos(ln

5. ∫ dxx

xx3

4

sin2

cos

- 6 -

解. ∫∫∫ ⋅==2

2sin

1

2sin

2cos

41

2cos

2sin8

2cos

sin2

cos

233

4

3

4

xdxx

xxdxxx

xxdx

x

xx

∫ ∫∫ +−=−=−= dxxxxxxdxdxx2

csc81

2csc

81

2csc

81

2csc

2csc

41 222

= cxxx +−−2

cot41

2csc

81 2

六. 求下列不定积分:

1. ∫ −++ dx

xxxx

22

2

)1()1ln(

解. ∫∫ −++=

−++

22

22

2

11)1ln(

21

)1()1ln(

xdxxdx

xxxx

= ∫+

⋅−

−−

++ dxxxx

xx222

2

11

11

21

11)1ln(

21

tx tan=令 tdtttx

xx 222

2

secsec

1tan11

21

)1(2)1ln(

⋅⋅−

−−

++∫

= dtt

tx

xx∫ −

−−

++22

2

sin21cos

21

)1(2)1ln(

= ∫ −−

−++

ttd

xxx

22

2

sin21sin2

221

)1(2)1ln(

= ctt

xxx

+−+

−−

++sin21sin21ln

241

)1(2)1ln(

2

2

= cxxxx

xxx

+−+

++−

−++

2121ln

241

)1(2)1ln(

2

2

2

2

2. ∫+

dxx

xx21

arctan

解. ∫∫∫ ++

−+=+=+

dxxxxxxxddx

xxx

2

222

2 11arctan11arctan

1arctan

= cxxxxdxx

xx +++−+=+

−+ ∫ )1ln(arctan11

1arctan1 22

2

2

- 7 -

3. ∫ dxe

ex

x

2arctan

解. dxe

eeeedeedxe

ex

xxxxxx

x

x

∫ ∫∫ ++−=−= −−−

2222

2 121arctan

21arctan

21arctan

dxe

eee x

xxx ∫ ++−=

−−

22

121arctan

21

∫ ++−= − dx

eeee xx

xx

)1(1

21arctan

21

22

cxeeedxe

ee

ee xxxx

x

xxx +++−=

+−+−= −−− ∫ )arctanarctan(

21)

11(

21arctan

21 2

22

七. 设

−+

−+=

−xexxxx

xf)32(

3)1ln()(

2

2

00

<≥

xx

, 求 ∫ dxxf )( .

解.

−+

−+=

−∫∫

∫ dxexx

dxxxdxxf

x)32(

)3)1ln(()(

2

2

+++−

+−+−−+=

−1

2

2222

)14(

3)]1ln([21)1ln(

21

cexx

cxxxxx

x

00

<≥

xx

考虑连续性, 所以 c =－1+ c1, c1 = 1 + c

∫ dxxf )(

++++−

+−+−−+=

− cexx

cxxxxx

x 1)14(

3)]1ln([21)1ln(

21

2

2222

00

<≥

xx

八. 设 xbxaef x cossin)(' += , (a, b为不同时为零的常数), 求 f(x).

解. 令 txet x ln== ， , )cos(ln)sin(ln)(' tbtatf += , 所以

∫ += dxxbxaxf )]cos(ln)sin(ln[)(

= cxabxbax+−++ )]cos(ln)()sin(ln)[(

2

九. 求下列不定积分:

1. ∫ +⋅+ dxxxx )32(3 32

解. ∫∫ +=+=+⋅+

++ cxddxxx

xxxx

3ln3)3(3)32(3

3233

222

- 8 -

2. ∫ −+− dxxxx )13()523( 23

2

解. ∫ −+− dxxxx )13()523( 23

2 = ∫ +−+− dxxxdxx )523()523(21 22

32

= cxx ++− 25

2 )523(51

3. ∫+

++2

2

1)1ln(

xxx

解. ∫∫ +++=++++=+

++ cxxxxdxxx

xx 2222

2

2

)]1[ln(21)1()1ln(

1)1ln(

4. ∫+++++ )11ln()11( 222 xxx

xdx

解. ∫∫++

++=

+++++ )11ln()11ln(

)11ln()11( 2

2

222 xxd

xxxxdx

= cx +++ |)11ln(|ln 2

十. 设当 x ≠ 0时, )(' xf 连续, 求 ∫+− dx

exxfxxxf

x2

)()1()('.

解. ∫ ∫∫ −−

=+− dx

xexfdx

exxfxxfdx

exxfxxxf

xxx)()()(')()1()('

22

= ∫ ∫−− dxxe

xfxxfde x

x )()(

=xxfe x )(− + ∫ dx

xexfx)(－ ∫ dx

xexfx)(

=xxfe x )(− +c.

十一. 设 xxxf 22 tansin)2(cos' +=+ , 求 f(x).

解.令 ，2cos += xt 2cos −= tx , 所以

22

22

)2(1)2(1

cos1cos1)('

−−−−=−+−=

tt

xxtf

所以 ∫ +−

−−−=

−

+−−= cx

xdxx

xxf2

1)2(31

)2(1)2()( 3

22

十二. 求下列不定积分:

1. ∫ +dx

xxx22 )1(

arctan

- 9 -

解. ∫∫ +−=

+ 222 11arctan

21

)1(arctan

xxddx

xxx

= ∫ ++

+− 222 )1(2

1)1(2

arctanx

dxx

x

tx tan=令 ∫++

− dttt

xx

4

2

2 secsec

21

)1(2arctan

= ∫++

− tdtx

x 22 cos

21

)1(2arctan

= cttx

x+++

+− 2sin

81

41

)1(2arctan

2

= cx

xxx

x+

+++

+−

)1(4arctan

41

)1(2arctan

22

2. ∫ +dx

xx

1arcsin

解. 令 u =x

x+1

, 2

2

1 uux−

= , duuudx 22 )1(

2−

=

∫∫ −=

+du

uuudx

xx

22 )1(2arcsin

1arcsin

tu sin=令 ∫ ∫ ∫== ttddtt

tttdtttt 2

34 tancossin2cos

cossin2

= ∫ ∫ ++−=−−=− cttttdtttttdttt tantan)1(sectantantan 22222

= cx

xxx

xx ++

+−+ 1

arcsin1

arcsin

3. ∫−

+⋅ dx

xx

xx

2

2

2 11arcsin

解. =−

+⋅∫ dx

xx

xx

2

2

2 11arcsin

∫−

dxxxx

22 1arcsin

+ ∫−

dxxx21

arcsin

tx sin=令 ∫ tdttt

t coscossin2 + ∫ +−= 22 )(arcsin

21cot)(arcsin

21 xttdx

= ∫ +++−=++− cxtttxtdttt 22 )(arcsin21|sin|lncot)(arcsin

21cotcot

= cxxxx

x+++

−− 2

2

)(arcsin21||lnarcsin1

- 10 -

4. ∫ +dx

xxx

)1(arctan

22

解. ∫∫∫∫ +−=

+−=

+− dx

xxdxxxdx

xxxdx

xxx

22

2222 1arctanarctan

111arctan

)1(arctan

= ∫ ∫−− − xxdxdx arctanarctanarctan 1

∫ −+

+−= − 22

1 )(arctan21

)1(1arctan xdx

xxxx

= ∫ −

+−+− − 2

21 )(arctan

21

111arctan xdxxx

xx

= cxx

xxx +−+

+− − 22

21 )(arctan

21

1ln

21arctan

十三. 求下列不定积分:

1. ∫ − dxxx 23 4

解. 令 tx sin2=

∫ ∫∫ −−==− ttdttdttdxxx coscos)cos1(32cossin324 222323

= cxxctt +−−−=++− 23

225

253 )4(34)4(

51cos

532cos

332

2. ∫ >− )0(

22

adxx

ax

解. 令 tax sec=

∫ ∫∫ +−===>− cattatdtatta

tataadx

xax tantantansec

sectan)0( 2

22

= cxaaax +−− arccos22

3. dxeee

x

xx

∫−

+21

)1(

解. =−

+∫ d

eee

x

xx

21)1(

∫−

dxe

ex

x

21+ dx

ee

x

x

∫− 2

2

1

= ∫− x

x

ede

21－ dx

eed

x

x

∫−

−2

2

1)1(

21

= cee xx +−− 21arcsin

- 11 -

4. ∫ −dx

xaxx

2

解. ∫ −dx

xaxx

2 xu =令 ∫

−du

uau

2

4

22 tau sin2=令 ∫ tdta 42 sin8

= ∫ ∫ +−=− dtttadtta )2cos2cos21(2

4)2cos1(8 22

22

= ctatatadttatata ++−=+

+− ∫ 4sin4

2sin232

4cos122sin224

22222

= ctttattata +−+− )sin21(cossincossin43 2222

= cttattata +−− cossin2cossin33 3222

= ca

xaax

axa

axa

axa

axa +

−−

−−

22

222

22

23

2arcsin3 222

= cxaxxaaxa +−

+− )2(

23

2arcsin3 2

十四. 求下列不定积分:

1. ∫ + xxdx

cos1sin

解. 令 ux =+ cos1 , )2(

21cos

2sin2

sin 2222 −=

−=

−=

uuudu

xudu

xudu

xdx

∫ ∫∫ −=−=

+ )2(2)2(

2

cos1sin 22

22

uududu

uuu

u

xxdx

= ∫ ∫∫ −+−=

−−−

2211

2222 udu

ududu

uu

= cuu

u+

−+

−22ln

2211

= cxx

x+

+−++

−+ cos12

cos12ln22

1cos11

2. ∫ +− dx

xx

cos2sin2

解. ∫∫∫ ++

++

=+−

xxddx

xdx

xx

cos2)cos2(

cos212

cos2sin2

- 12 -

tx=

2tan令 ∫∫ ++

+=++

+−

+

+ |cos2|ln322|cos2|ln

112

12

2 2

2

2

2x

tdtx

tt

tdt

= cxxcxt+++=+++ |cos2|ln)

2(tan

31arctan

34|cos2|ln

3arctan

34

3. ∫ +dx

xxxx

cossincossin

解. ∫∫ +−+

=+

dxxxxxdx

xxxx

cossin1cossin21

21

cossincossin

= ∫ ∫ ∫ +−+=

+−+ dx

xxdxxxdx

xxx

cossin1

21)cos(sin

21

cossin1cos)(sin

21 2

= ∫+

+−−

)4

sin(

)4

(

42)cos(sin

21

π

π

x

xdxx

= cxxx ++−− |)82

tan(|ln42)cos(sin

21 π

十五. 求下列不定积分:

1. ∫ −dx

xxx

1

解. cxxx

xddxx

xdxxx

x+−−=

−

−=

−=

− ∫ ∫∫ 134

)1(

)1(32

)1(1 21

23

23

21

23

21

2. ∫ +− dx

ee

x

x

11

解. =+−

∫ dxee

x

x

11

∫ −

−

+− dx

ee

x

x

11

= =−

−∫ −

−

dxee

x

x

211

∫ −−dx

e x211

+ ∫ −

−

− x

x

ede

21

= ∫ −+−

x

x

x

ee

dxe arcsin12

te x sec=令 ∫ −+ xettdtt arcsin

tantansec

= cettetdt xx +++=+∫ −− arcsin|sectan|lnarcsinsec

= ceee xxx ++−+ −arcsin|1|ln 2

- 13 -

3. ∫−− dx

xxx 1arctan1

解. ∫−− dx

xxx 1arctan1

= ∫ −+−

−− dxxx

xx 121)1(

1arctan12

ux =−1令 ∫ ∫ ∫ +−=

+du

uuududu

uuu

1arctan2arctan2

1arctan2

22

2

= ∫ −+

− 22 )(arctan

12arctan2 udu

uuuu

= cuuuu +−+− 22 )(arctan)1ln(arctan2

= cxxxx +−−−−− 2)1(arctan||ln1arctan12

- 1 -

- 1 -

- 2 -