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- 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
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