3a4xa2xa2?x2222?arcsin?xa?x?(a?2x2)?C.
8a28 ***8.
?x2?6x?5dx.
x?3解：原式??(x?3)2?4dx(令x?3?2sect)
x?32tant?2sect?tant?dt?2?(sec2t?1)dt?2tant?2t?C 2sect2x2?6x?5?2arccos?C.
x?3?? ? ***9.
?dx15?2x?x2.
解:
?dx15?2x?x2??.
d(x?1)16?(x?1)2?arcsinx?1?C． 4**10.
?2dx(x?1)(x?3)解：令x?2?sect，
?dsectsect?1??sectdt?lnsect?tant?C1
?ln(x?2)? ***11.
x2?4x?3?C1?2ln(x?3?x?1)?C.
?dx1?ex.
解: 设1?ex?t. 则ex?t2?1 exdx?2tdt
?原式??2tdtt?1dt?2?ln?t2?1t?1?C t(t2?1)t2?1x ?ln?c?x?2ln1?1?e?C. 2(t?1)
?? 74
第6章 （之3） 第27次作业
教学内容：§6.1.3不定积分的分部积分法 **1.
?lnxxdx.
解:
?lnx1dx?lnxd(2x)?2x?lnx??2x?dx
xx? ?2xlnx?2
**2. xsinxcosxdx.
?1dx ?2xlnx?4x?C . x?11x?sin2x?dx??x?dcos2x 2?4?1111 ??xcos2x??cos2x?dx??xcos2x?sin2x?C.
4448解：原式? **3.
xsinx?cos3xdx.
xdcosx1dx?xd(cosx)?2 ?cos3x?21111x1?2dx??tanx?C. ?x(cosx)??2222cosx2cosx2解：原式??
**4.x?tanx?secx?dx.
?41x?dsec4x ?411x1?x?sec4x??sec4x?dx?sec4x??(tan2x?1)dtanx 4444x114tan3x?tanx?C. ?secx?4124解：原式?3?x?secx?dsecx?
x2***5. ecosxdx.
? 75
x22xx2x解： ecosxdx?cosx?de?ecosx?2ecosxsinxdx
????excos2x??sin2x?dex
xxx sin2x?de?esin2x?2ecos2xdx
???exsin2x?2?cos2xdex?exsin2x?2excos2x?4?exsin2xdx
1x2esin2x?excos2x?C 551x2xx2 原式?ecosx?esin2x?ecos2x?C.
551?cos2x2注：也可先将 cosx 写成 .
21x1x1ecos2x?sin2x?C 答案也可以是 ：e?2105?**6. ln(cosx)?cos2x?dx. 解： ln(cosx)?cos2x?dx ??1ln(cosx)?dsin2x ?2?11?sinx11?cos2x?sin2x?ln(cosx)??sin2x?dx?sin2x?ln(cosx)???dx 22cosx22111?sin2x?ln(cosx)?x?sin2x?C. 224
**7. sinx?lntanxdx. 解: 原式??lntanxd(cosx)
????cosx?lntanx??cosx???cosxlntanx??
**8.
1?sec2xdx tanx1xdx??cosx?lntanx?lntan?C. sinx2x?cosx?sin2xdx.
解: 原式??xd( ?? **9.
?1x1)????dx sinxsinxsinx
x?lncscx?cotx?C. sinx?arctanxdx.
x2(1?x2)76
arctanxarctanxdx??x2?1?x2dx 1???arctanxd??arctanxdarctanx
x1111??arctanx???dx?arctan2x 2xx1?x211122 ??arctanx?lnx?ln1?x?arctanx?C.
x22解：原式? **10.
?xarcsinx1?x2dx.
解：
?xarcsinx1?x2dx????x1?x22arcsinxdx??arcsinxd1?x2
dx1?x2? ??1?x?arcsinx??1?x2?
??1?x2arcsinx?x?C.
**11. xsin解: 令?xdx.
x?u,则x?u2， dx?2udu，
原式?2?u3sinudu?2?u3cosu?3?u2cosudu ?2(?ucosu?3usinu?6?usinudu)3232??
?2(?ucosu?3usinu?6ucosu?6sinu)?C ?2u(6?u)cosu?6(u?2)sinu?C ?2x(6?x)cosx?6(x?2)sinx?C. ***12.
22?xexe?2xdx.
解：
?xexex?2dx(令t?ex?2,x?ln(t2?2))
(t2?2)?ln(t2?2)2t??2?dt
tt?2? 77
?2ln(t2?2)dt?2t?ln(t2?2)?2t??2t?ln(t2?2)?4(1???2t?dtt2?2
?2)dt2t?2t2?C?2t?ln(t2?2)?4t?42arctanex?2 ?2x?e?2?4e?2?42arctan?C.
2xx**13.e?arcsinxdx.
arcsinx解：原式?x?e??x?earcsinx?11?x2dx
?x?earcsinx??earcsinxd1?x2
?x?earcsinx?1?x2?earcsinx??1?x2?earcsinx? ?
x**14. 已知 f?(e)?x,f(1)?0，求f(x).
11?x2?dx
1(x?1?x2)earcsinx?C. 2df(ex)xx?x解一：已知 f?(e)?x,即， 或 ， df(e)?x?dexdex 两边积分，得 f(e)?xe?e?C， 由 f(1)?0， 得 C?1，
xxx 故f(e)?xe?e?1 令e?u，得
xxxx f(u)?u?lnu?u?1， 即 f(x)?x?lnx?x?1。
xx解二：已知f?(e)?x，令e?u，则有
f?(u)?ln(u)， 两边积分，得 f(u)?u?lnu?u?C，
由f(1)?0，得C?1.
所以f(u)?u?lnu?u?1，即 f(x)?x?lnx?x?1。
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