1、求下列不定积分 1)
?xsin2xdx;
x?1?cos2x??xxcos2x?dx???dx ???222??解:原式?x2xsin2xsin2xx2xsin2xcos2x???dx????C ?4444482)
??1?ln?1??dx; ?x???1?x11?x??dx?dx?lnx?1?C ??2?x?x?1xx?1解:原式?xln?1?3)
?arctanxdx;
解:原式?xarctanx?x?1?1?2x?1?x?dx
1?11 ?xarctanx??x2dx??dx
22x?1?x?1?3 ?xarctanx?x2?arctanx?C
44)
?xtan2xdx;
2x2?xtanx??tanxdx 解:原式??x?secx?1?dx??2x2?xtanx?lncosx?C ??25)
?arcsinxdx; 2x?解:令x?sint原式???2?t??2
arcsinx1arcsinx1??dx????dt
2xxsintx1?xarcsinxarcsinx11?x2?lncsct?cott?C???ln??C ??xxxx6)
?x3e?xdx;
2x2?x2x2?x21?x2?x2e??xedx??e?e?C 解: 原式??2227)
?sinxlntanxdx;
sec2xdx??cosxlntanx??cscxdx 解:原式??cosxlntanx??cosxtanx ??cosxlntanx?lncscx?cotx?C 8)
?xarctanx1?x2dx;
11?x2解:原式?1?xarctanx?2?dx
?1?x2arctanx?lnx?1?x2?C
??9)
?xexe?1xdx;
?ex1?e?1dx?2xe?1?2????dx xxe?1??e?1xxx2x?2解:原式?2xe?1?2
x??2xe?1?4e?1?2?xx?e??1??e????2dx?2xe?1?4e?1?2arcsinexx?x2?C
10)xlnx?1?x2dx;
???解:方法一、令x?tant
x2x22原式?lnx?1?x??dx
2221?x??x2tan2t2lnx?1?x???sec2tdt ?22sect??x21lnx?1?x2??tan2t?sectdt ?22?? 因为tan2t?sectdt?tantsect?sec3tdt?tantsect?????sect?tan2tsect?dt
?tantsect?lnsect?tant??tan2tsectdt
112tantsectdt?tantsect?lnsect?tant?C ?22x211原式?lnx?1?x2?tantsect?lnsect?tant?C
244x211 ?lnx?1?x2?x1?x2?lnx?1?x2?C
244方法二、
???? 设u?x??lnx?1?x?2?x21,v??x??x,则v?x???
24?x21??x21?12dx 原式????lnx?1?x?????22424????1?x???x21?12x2?12dx ????lnx?1?x??22441?x?????x21?1?x222 ????lnx?1?x???1?x?4?1?x2?24?????dx ??x21?1 ????lnx?1?x2?x1?x2?C
4?24???2、已知f(x)的一个原函数是
sinx,求?xf?(x)dx; x解:
?sinx??sinx?xf?(x)dx?xf?x???f?x?dx?x??C ??xx?? ?xcosx?2sinx?C
xx3、设f?(e)?1?x,求f(x).
解:令t?e,则x?lnt,f??t??1?lnt,f?t??x?f??t?dt???1?lnt?dt?tlnt?C
所以f?x??xlnx?C
作业22不定积分的分部积分法
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