131342tanxdx?tanx?tanxdx?tanx?tanx??dx??33（2）
1?tan3x?tanx?x?C3cos3xsin3x3??cos2xsin2xdx （3）?cosxsinxdx??6624cos3xsin3x1????sin22xdx
68cos3xsin3x11?cos4x????dx
682cos3xsin3xx1????sin4x?C
616646．导出下列不定积分对于正整数n的递推公式：
(1)?n??xnekxdx; (2)?n??(lnx)ndx;
(3)?n??(arcsinx)ndx;(4)?n??eaxsinnxdx;解：（1）?n?
1nkx1nkxnn?1kxxde?xe??xedx k?kk1n?xnekx??n?1. kknn?11（2）?n?x(lnx)?n?x(lnx)?dx
x?x(lnx)n?n?n?1.
（3）?n?x(arcsinx)?nn?x1?x2(arcsinx)n?1dx
?x(arcsinx)n?n?(arcsinx)n?1d1?x2
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?x(arcsinx)n?n1?x2(arcsinx)n?1?n(n?1)??arcsinx??x?arcsinx??n1?x2?arcsinx?（4）?n?nn?1n?2dx
?n?n?1?In?2
11axnnaxn?1naxsinxde?esinx?esinxcosxdx ??aaa1nnn?22nax??eaxsinnx?2?eaxsinn?1xcosx?2??n?1sinxcosx?sinxe???dx aaa?1nn?eaxsinnx?2?eaxsinn?1xcosx?2??eaxdx ?n?1?sinn?2x?nsinnx???aaan?n?1?1axn?1n2?2esinx?asinx?ncosx??In?2?2In aa2a 移项合并得
I?1?eaxsinn?1x?asinx?ncosx??n?n?1?In?2?22??. n?a7．利用上题所得递推公式计算：
(1)?x3e2xdx;
(2)?(lnx)3dx;(3)?(arcsinx)3dx;(4)?exsin3xdx.32x
解：（1）xedx?132x32xe??xe2xdx ?2213?1??x3e2x??x2e2x??xe2xdx?22?2?132x322x3?12x12x?xe?xe??xe??edx?242?22? 1333?x3e2x?x2e2x?xe2x?e2x?C2448333??1?e2x?x3?x2?x???C448??2?（2）(lnx)dx?x?lnx??3?33??lnx?dx
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2
32?x?lnx??3?x?lnx??2?lnxdx??? ?x?lnx??3x?lnx??6?xlnx?x??C
32?x??lnx??3x?lnx??6xlnx?6??C??3（3）(arcsinx)dx
32??x(arcsinx)3?31?x2?arcsinx??6arcsinxdx?x(arcsinx)3?31?x2?arcsinx??6xarcsinx?6?222xdx1?x2
?x(arcsinx)3?31?x2?arcsinx??6xarcsinx?61?x2?C（4）esinxdx??x31?x2xesinxsinx?3cosx?6esinxdx???????, 10xxxx而esinxdx?sinxde?esinx?ecosxdx
????exsinx??cosxdex?esinx?ecosx??esinxdx,移项解得
x?esinxdx?xxx
1xe?sinx?cosx??C1, 21x23x?esinxsinx?3cosx?3e故有?(arcsinx)dx? ???sinx?cosx??C1???101?ex?sin3x?3sin2xcosx?3sinx?3cosx??C. 10
§3 有理函数和可化为有理函数的不定积分
1．求下列不定积分：
x3x3?1?11dx??dx??(x2?x?1?)dx ⑴ ?x?1x?1x?1x3x2???x?ln|x?1|?C 32
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x?221(x?4)2⑵ 解法一：?2dx??(?)dx?ln?C
x?4x?3|x?3|x?7x?12解法二：
x?212x?713dx?dx?dx 22?x2?7x?12??2x?7x?122x?7x?121d(x2?7x?12)3??2??22x?7x?1217d(x?)
712(x?)2?24?13x?4ln|x2?7x?12|?ln?C 22x?311ABx?C ???3221?x(1?x)(1?x?x)1?x1?x?x⑶ 解
去分母得 1?A(1?x?x2)?(Bx?C)(1?x)
令x??1，得A?13. 再令x?0，得A?C?1，于是C?23. 比较上式两端二次幂的系数得 A?B?0，从而B??13，因此
dx1dx1x?2???1?x33?1?x3?1?x?x2dx112x?111?ln|1?x|??dx?dx22?361?x?x21?x?x1111?ln|1?x|?ln(1?x?x2)??dx2362(x?12)?341(1?x)212x?1?ln?arctan?C 61?x?x233dx1(1?x2)?(x2?1)11?x21x2?1??dx??dx??dx ⑷ 解 ?4444221?x21?x1?x1?x11111?d(x?)d(x?)221111xxxx ??dx??dx????1111222222x?2x?2x2?2x2?2xxxx1?
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11d(x?)d(x?)1x?1x ???1122(x?)2?2(x?)2?2xx122x?arctan11x??2x?1lnx?C
1242x??2x?2x2?12x2?2x?1?arctan?ln2?C 482xx?2x?1⑸
dx?(x?1)(x2?1)2
解 令
1ABx?CDx?E, 解得 ???22222x?1x?1(x?1)(x?1)(x?1)A?111, B?C??, D?E??, 于是 442dx1dx1x?11x?1??dx??(x?1)(x2?1)24?x?14?x2?12?(x2?1)2dx ?111111xln|x?1|?ln(x2?1)?arctanx??(arctanx?)?C 224844x?14x?1?1|x?1|1?x(ln?2arctanx?2)?C 4x?1x2?1⑹
x?214x?251dx?dx?dx 2222?(2x2?2x?1)2??4(2x?2x?1)2(2x?2x?1)4x?2d(2x2?2x?1)1其中? dx???22222?(2x?2x?1)(2x?2x?1)2x?2x?1141dx?dx?2?(2x2?2x?1)2?[(2x?1)2?1]2?[(2x?1)2?1]2d(2x?1)
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