??1x11dx?dx?dx?lnx?d(x2?1)222???x2x?1x(x?1)x?11x1?lnx?ln(x2?1)?C?ln?C.22x?1★★★(12)
dx?(x2?x)(x2?1)
思路：将被积函数裂项后分项积分。 解：?11 ?(x2?x)(x2?1)x(x?1)(x2?1)令
1ABCx?D，等式右边通分后比较两边分子x的同次项的系数得： ???222(x?x)(x?1)xx?1x?1A?B?C?0,A?C?D?0,A?B?D?0,A?1，解之得：
111A?1,B??,C??,D??.
22211111x?1?2?????(x?x)(x2?1)x2x?12x2?111111x11 ???????2222x2x?122(x?x)(x?1)x?1x?1dx1111x1dx??2?dx?dx?dx?2?x?12?x2?12?x2?1(x?x)(x2?1)?x?1111?lnx?lnx?1??2d(x2?1)?arctanx24x?12
111?lnx?lnx?1?ln(x2?1)?arctanx?C.242★★★★★(13)
dx?x4?1
思路：将被积函数裂项后分项积分。
解：?x4?1?(x2?1?2x)(x2?1?2x)
令
1Ax?BCx?D??，等式右边通分后比较两边分子x的同次项的系数得：
x4?1x2?1?2xx2?1?2x 36
?2A???4?A?C?0?1??B????2A?B?2C?D?02解之得：???C?2?A?2B?C?2D?0??4B?D?1???D?1??2?
112x?212x?22(2x?2)?22(2x?2)?2??????4x2?1?2x4x2?1?2x88x4?1221221(x?)?(x?)?22222(2x?2)(2x?2)111?[?]?[?]84221221221221(x?)?(x?)?(x?)?(x?)?22222222dx2(2x?2)(2x?2)111??4?[?]dx?[?]dx?8?4x?1221221221221(x?)?(x?)?(x?)?(x?)?22222222?2(2x?2)(2x?2)1[?2dx??2dx]?[?84x?1?2xx?1?2x1(x?221)?22dx??1(x?221)?22dx]?211[?2d(x2?1?2x)??2d(x2?1?2x)]8x?1?2xx?1?2x?211[?d(2x?1)??(2x?1)2?1d(2x?1)] 4(2x?1)2?12x2?2x?12?ln2?[arctan(2x?1)?arctan(2x?1)]?C
8x?2x?142x2?2x?122x??ln2?(arctan)?C. 281?xx?2x?14注：由导数的性质可证arctan(2x?1)?arctan(2x?1)?arctan本题的另一种解法：
2x1?x2
37
11x2?1x2?1?4?[4?4]x?12x?1x?1111?22dx1x?1x?11xx??4?[dx??4dx]?[?dx??dx]112x?12?x4?1x?1x2?2x2?2xx11111?[?d(x?)??d(x?)]112xxx2?2x2?2xx11111?[?d(x?)??d(x?)]112xx(x?)2?2(x?)2?2xx1x?21111x)?2[(?d(?)d(x?)]??11148x2x?(x?)?2(x?)?2
2xxx()?12221?2?4?1x2?1211d()?[d(x??2)2?18xx?122xx??21?()x2x 1x??2112x2?12x??d(x??2)]?arctan?ln?C11x482xx??2x??2xx2x2?12x2?2x?1?arctan?ln2?C 482xx?2x?12x2?2x?122x?ln2?(arctan)?C. 281?xx?2x?14注：由导数的性质可证arctanx2?12x??2?arctan2x。 21?x?x2?2★★★★★(14)?(x2?x?1)2dx
思路：将被积函数裂项后分项积分。
?x2?2x2?x?1?x?1解：?2 ??(x?x?1)2(x2?x?1)2??112x?131??x2?x?12(x2?x?1)22(x2?x?1)2
38
?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)?dx2222???2x?132(x?x?1)2(x?x?1)()2?13??232x?11131arctan()??dx222?32x?x?12(x?x?1)3
又?3112x?1dx dx??2?(x2?x?1)22x2?x?1?x2?x?112x?1232x?1?arctan()?C2x2?x?133?x?2432x?1x?1dx??arctan()??C.23(x2?x?1)2x?x?132???
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??dx(x2?a2)n，其中n为正整数，a?0，则必有：
In?1x[?(2n?3)In?1]。
2a2(n?1)(x2?a2)n?12、 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 39
??★★(2)
3332arctan(cotx)?C?arctan(tanx)?C. 6263dx?3?cosx
思路：万能代换！
x1?t22dt,dx?; 解：令t?tan，则cosx?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,dx?; ，则sinx?21?t21?t22t?12dtd()2dxdtdt23????1?t??2????2t?122t1232?sinxt?t?132?(t?)?1?()2241?t322t?1?arctan()?C33dx2?arctan(2?sinx32tanx?12)?C. 3
??dx?1?tanx
思路：利用变换t?tanx！（万能代换也可，但较繁！）
★★(4)
40