?x2?2dx12x?131??2dx???dx?dx222222???2(x?x?1)(x?x?1)x?x?12(x?x?1)dx11312?????2d(x?x?1)?dx22?1232(x?x?1)22(x?x?1)(x?)?24dx11312?????2d(x?x?1)?dx222?2(x?x?1)2(x?x?1)13(x?)2?()2222x?1d()23113132???d(x?x?1)?dx3?2x?122?(x2?x?1)22?(x2?x?1)2()?13??232x?11131arctan()??dx222?32x?x?12(x?x?1)3
又?3112x?1dxdx??
2?(x2?x?1)22x2?x?1?x2?x?1???12x?1232x?1?arctan()?C2x2?x?133?x?2432x?1x?1dx??arctan()??C.23(x2?x?1)2x?x?13dx(x2?a2)n2
注：本题再推到过程中用到如下性质：（本性质可由分部积分法导出。）
若记 In??，其中n为正整数，a?0，则必有：
In?1x[?(2n?3)In?1]。 222n?12a(n?1)(x?a)2、 求下列不定积分
知识点：三角有理函数积分和简单的无理函数积分法的练习。
思路分析：求这两种积分的基本思路都是通过适当的变换化为有理函数积分去完成。
★★(1)
dx?3?sin2x
22思路：分子分母同除以sinx变为cscx后凑微分。
dxcsc2xdxdcotx3?????解：??3cot2x?46?3?sin2x?3csc2x?1d((3cotx)2?123cotx)2 36
??★★(2)
3332arctan(cotx)?C?arctan(tanx)?C. 6263dx?3?cosx
思路：万能代换！
1?t22dtx解：令t?tan，则cosx?,dx?; 2221?t1?t2dt2dxdt1t1?t??????arctan?C22?1?t3?cosx2?t223?
1?t2dx11x???arctan(tan)?C.3?cosx222注：另一种解法是：
xdxdx1dx12dx ????3?cosx???xx2x3?2cos2?121?cos2sec2?12221x1x11x ?dtan??dtan?arctan(tan)?C.
xx22222tan2?2(tan)2?(2)222dx★★(3)?2?sinx
sec2思路：万能代换！ 解：令t?tanx2t2dt，则sinx?,dx?; 21?t21?t22t?12dtd()2dxdtdt23????1?t??2????2t?122t1232?sinxt?t?132?(t?)?1?()2241?t322t?1?arctan()?C33
x2tan?1dx22)?C. ???arctan(2?sinx33dx?1?tanx
思路：利用变换t?tanx！（万能代换也可，但较繁！）
★★(4)
37
解：令t?tanx，则x?arctant,dx?dt; 1?t2dt2dxdt ????1?t??21?tanx1?t(1?t)(1?t)?111t?111t1?(?)?(??)2222(1?t)(1?t)21?t1?t21?t1?t1?tdt11t1???(dt?dt?dt)222???(1?t)(1?t)21?t1?t1?t
11?[ln1?t?ln(1?t2)?arctant]?C22dx11???[ln1?tanx?ln(1?tan2x)?x]?C.1?tanx22★★(5)
dx?1?sinx?cosx
思路：万能代换！
2t1?t22dtx解：令t?tan，则sinx?,cosx?,dx?; 22221?t1?t1?t2dt2dtx1?t????ln1?t?C?ln1?tan?C 2?2t1?t1?t21??1?t21?t2dx★★(6)?5?2sinx?cosx
思路：万能代换！
2t1?t22dtx,cosx?,dx?; 解：令t?tan，则sinx?21?t21?t21?t22dtdxdt1?t2 ?????22?5?2sinx?cosx2t1?t3t?2t?25?2?1?t21?t2d(3t?15)?dt1??而?23t?2t?23dt15(t?)2?()233?1?5153t?12()?15arctan(3t?15)?C
x3tan?1dx12)?C. ???arctan(5?2sinx?cosx55 38
★★★★(7)
dx?(5?4sinx)cosx
思路一：万能代换！
2t1?t22dtx解：令t?tan，则sinx?,cosx?,dx?; 22221?t1?t1?t2dt2dx2(1?t2)dt1?t???22(5?4sinx)cosx2t1?t(5t?8t?5)(1?t2)(5?4)
1?t21?t224??(2?2)dt5t?8t?5(5t?8t?5)(t2?1)而
44?，
(5t2?8t?5)(t2?1)(5t2?8t?5)(t?1)(t?1)4At?BCD???，等式右边通分后比较两边分子t的同22(5t?8t?5)(t?1)(t?1)5t?8t?5t?1t?1令
次项的系数得：
?A?5C?5D?0?5A=?B?13C?3D?0???2,解之得：???A?13C?3D?0??B=7???8?B?5C?5D?41?C???16; ??D??9?16??4120t?71191?????? 22(5t?8t?5)(t?1)(t?1)85t?8t?516t?116t?11191110t?891?????2??216t?116t?145t?8t?585t?8t?5dx1191110t?871??(??????2??2)dt(5?4sinx)cosx16t?116t?145t?8t?585t?8t?5dx1191110t?871?????dt??dt??2dt??2dt(5?4sinx)cosx16t?116t?145t?8t?585t?8t?519175t?4??lnt?1?lnt?1?ln(5t2?8t?5)?arctan()?C16164243x5tan?41x9x1xx72??lntan?1?lntan?1?ln(5tan2?8tan?5)?arctan()?C162162422243?思路二：利用代换t?sinx！ 解：令t?sinx，x
39
dt2dxdtdt1?t?????????(5?4t)(t2?1)
(5?4sinx)cosx(5?4t)(1?t2)(5?4t)1?t211??(5?4t)(t2?1)(5?4t)(t?1)(t?1)令
1ABC，等式右边通分后比较两边分子t的同次项的系数得： ???2(5?4t)(t?1)5?4tt?1t?116?A??9?A?4B?4C?0?111611111????????? ?9B?C?0解之得:?B?218(5?4t)(t?1)95?4t18t?12t?1??A?5B?5C?1??1?C???2???dt1611111?dt?dt?dt(5?4t)(t2?1)9?5?4t18?t?12?t?1
411?ln5?4t?ln1?t?ln1?t?C9182??dx411??ln5?4sinx?ln1?sinx?ln1?sinx?C.
(5?4sinx)cosx9182注：比较上述两解法可以看出应用万能代换对某些题目可能并不简单！
★★★★(8)
1?sinx?(1?cosx)sinxdx
思路：将被积函数分项得，对两个不定积分分别利用代换t?cosx和万能代换！ 解：?1?sinx11 ??(1?cosx)sinx(1?cosx)sinx1?cosx??1?sinx11dx??dx??dx
(1?cosx)sinx(1?cosx)sinx1?cosxdt1dx??,sinx?1?t2; dx，令，则t?cosx,x?(0,?)?(1?cosx)sinx1?t2?dt对积分
1dtdt1?t2??dx????? 22?2(1?cosx)sinx(1?t)(t?1)(1?t)(t?1)(1?t)1?t令
1ABC???(1?t)2(t?1)t?11?t(1?t)2，等式右边通分后比较两边分子t的同次项的系数得：
40
不定积分例题及答案 理工类 吴赣昌
