?3arctanx?12x?1?ln?C
23x?x?11?x?x2★★★(8)?(x2?1)2dx
思路：将被积函数裂项后分项积分。
1?x?x21x2????解：?2(x?1)2x2?1(x2?1)2(x2?1)2
1?x?x21xdx??2dx??dx?dx?2?x2?1?(x2?1)2?(x2?1)2(x?1)2111dx2???2dx??2d(x?1)?2?(x2?1)22(x?1)2x?1又由分部积分法可知：2
dxx1???(x2?1)2x2?1?x2?1dx
1?x?x2x1112x?1??2dx???C?()?C
(x?1)2x2?12x2?12x2?1★★★(9)
xdx?(x?1)(x?2)(x?3)
思路：将被积函数裂项后分项积分。 解：?xx?3?313 ???(x?1)(x?2)(x?3)(x?1)(x?2)(x?3)(x?1)(x?2)(x?1)(x?2)(x?3)令
3ABC， ???(x?1)(x?2)(x?3)x?1x?2x?3等式右边通分后比较两边分子x的同次项的系数得：
3?A?33??A?B?C?02?33?5A?4B?3C?0?2??2 解之得：??B??3?(x?1)(x?2)(x?3)x?1x?2x?3?6A?3B?2C?3?3??C?2?而
111??
(x?1)(x?2)x?1x?2 31
3x112?????2(x?1)(x?2)(x?3)2x?1x?2x?3xdx11dx3dx ?????dx?2???(x?1)(x?2)(x?3)2x?1x?22x?313??lnx?1?2lnx?2?lnx?3?C.22x2?1★★★(10)?(x?1)2(x?1)dx
思路：将被积函数裂项后分项积分。
x2?1x2?1?212???解：?
(x?1)2(x?1)(x?1)2(x?1)x?1(x?1)2(x?1)令
2ABC???，等式右边通分后比较两边分子x的同次项的系数得：
(x?1)2(x?1)x?1x?1(x?1)2A?B?0,2A?C?0,11A?B?C?2；解之得：A?,B??,C??1。
221122?2?1??(x?1)2(x?1)x?1x?1(x?1)211x?12?2?1??(x?1)2(x?1)x?1x?1(x?1)22
x2?11dx1dx1??dx???dx 22???(x?1)(x?1)2x?12x?1(x?1) ?11111lnx?1?lnx?1??C ?lnx2?1??C. 22x?12x?1★★★(11)
?x(x112?1)dx
思路：将被积函数裂项后分项积分。 解：令
x(x2?1)?ABx?C?2，等式右边通分后比较两边分子x的同次项的系数得： xx?1?A?B?0?A?111x??C?0B??1???解之得： ??22x(x?1)xx?1?A?1?C?0?? 32
??1x11dx?dx?dx?lnx?d(x2?1)222???x2x?1x(x?1)x?1x11?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?2??????? (x?x)(x2?1)x2x?12x2?12x2?1dx1111x1dx??2?dx?dx?dx?2?x?12?x2?12?x2?1(x?x)(x2?1)?x1111?lnx?lnx?1??2d(x2?1)?arctanx24x?12
111?lnx?lnx?1?ln(x2?1)?arctanx?C.242★★★★★(13)
dx?x4?1
22思路：将被积函数裂项后分项积分。 解：?x?1?(x?1?2x)(x?1?42x)
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 33
?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?2x211[?d(2x?1)?d(2x?1)] ?224(2x?1)?1(2x?1)?1?2x2?2x?12?ln2?[arctan(2x?1)?arctan(2x?1)]?C
8x?2x?142x2?2x?122x??ln2?(arctan)?C. 2841?xx?2x?1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
34
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?1?(1x2?12xd()2x2?12x)?2[?811d(x??2)1xx??2x 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?arctan2x1?x2。
?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
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不定积分例题及答案 理工类 吴赣昌
