中国人民大学出版社（第四版）高等数学一第4章课后习题详解(7)

又?x?f(f?1(x))

??f?1(x)dx?f?1(x)??xd(f?1(x))?f?1(x)??f(f?1(x))d(f?1(x))

又??f(x)dx?F(x)?C

??f?1(x)dx?f?1(x)??f(f?1(x))d(f?1(x))?f?1(x)?F(f?1(x))?C.

习题4-4

1、 求下列不定积分

知识点：有理函数积分法的练习。

思路分析：被积函数为有理函数的形式时，要区分被积函数为有理真分式还是有理假分式，若是假分式，

通常将被积函数分解为一个整式加上一个真分式的形式，然后再具体问题具体分析。

x3★(1)?x?3dx

思路：被积函数为假分式，先将被积函数分解为一个整式加上一个真分式的形式，然后分项积分。

x3x3?27?2727解：? ??x2?3x?9?x?3x?3x?3x32727??dx??(x2?3x?9?)dx??(x2?3x?9)dx??dxx?3x?3x?3 13?x3?x2?9x?27lnx?3?C.32x5?x4?8★★★(2) ?x3?xdx

思路：被积函数为假分式，先将被积函数分解为一个整式加上一个真分式的形式，然后分项积分。

x5?x4?8(x5?x3)?(x4?x2)?(x3?x)?x2?x?8x2?x?82??x?x?1?3, 解：?33x?xx?xx?x而x3?x?x(x?1)(x?1),

x2?x?8ABC???令，等式右边通分后比较两边分子x的同次项的系数得：

x3?xxx?1x?1?A?B?C?1?A?8???C?B?1解此方程组得：?B??4 ??C??3A?8?? 26

x5?x4?88432??x?x?1???xx?1x?1x3?xx5?x4?88432??dx?(x?x?1???)dx ?xx?1x?1x3?x11?x3?x2?x?8lnx?4lnx?1?3lnx?1?C32★★★(3)

3?x3?1dx

2思路：将被积函数裂项后分项积分。 解：?x?1?(x?1)(x?x?1)，令

33ABx?C等式右边通分后比较两边分子??32x?1x?1x?x?1x的同次项的系数得：

?A+B=0?A?1???B+C-A=0解此方程组得：?B??1?A+C=3?C?2??

13(2x?1)?31?x?212?3??2??2x?1x?1x?x?1x?113(x?)2?()222 1(2x?1)1312???x?1(x?1)2?3213(x?)2?()224221(2x?1)31312??3dx??dx??dx??dx123x?1x?1213(x?)?(x?)2?()224221111312)?lnx?1??d((x?)2?)?3?d(12(x?1)2?3243x?242)2?12(3212x?1?lnx?1?ln(x2?x?1)?3arctan()?C.23x?★★

(4)

x?1?(x?1)3dx

思路：将被积函数裂项后分项积分。

27

解：令

得：

x?1ABC???(x?1)3x?1(x?1)2(x?1)3，等式右边通分后比较两边分子

x的同次项的系数

A?0,B?2A?1,?A?B?C?1，解此方程组得：A?0,B?1,C?2。

x?112??(x?1)3(x?1)2(x?1)3

x?11211x??dx?dx?dx????C???C?(x?1)2?(x?1)3x?1(x?1)2(x?1)3(x?1)2★★★(5)

3x?2?x(x?1)3dx

思路：将被积函数裂项后分项积分。 解：?3x?2322ABCD??????，令

x(x?1)3(x?1)3x(x?1)3x(x?1)3xx?1(x?1)2(x?1)3

等式右边通分后比较两边分子x的同次项的系数得：

A?B?0??A?2?3A?2B?C?0?B??2??解此方程组得：???3A?B?C?D?0?C??2??A?2??D??2。

?22222????x(x?1)3xx?1(x?1)2(x?1)3

3x?2322221222?????????x(x?1)3(x?1)3xx?1(x?1)2(x?1)3(x?1)3xx?1(x?1)23x?21222??dx?dx?dx?dx??(x?1)3?(x?1)2?x?1?xdxx(x?1)3112????2lnx?1?2lnx?C2(x?1)2x?1??2ln★★★(6)

x4x?3??C.2x?12(x?1)xdx?(x?2)(x?3)2

思路：将被积函数裂项后分项积分。 解：?xx?2?2x?22??? 2222(x?2)(x?3)(x?2)(x?3)(x?2)(x?3)(x?2)(x?3) 28

?122ABC；令，等式右边通????2222x?2x?3(x?3)(x?3)(x?2)(x?3)(x?2)(x?3)分后比较两边分子x的同次项的系数得：

A?B?0??A?22222????? ?6A?5B?C?0解此方程组得：?B??2?22x?2x?3(x?2)(x?3)(x?3)?9A?6B?2C?2?C??2??x1222322??(??)???(x?2)(x?3)2(x?3)2x?2x?3(x?3)2(x?3)2x?2x?3xdx322???dx?dx?dx 22???(x?2)(x?3)(x?3)x?2x?3?33?x?3????2lnx?2?2lnx?3?C?ln???C.?x?3?x?2?x?3★★★(7)

23x?x3?1dx

思路：将被积函数裂项后分项积分。

3x3(x?1)?333 ???3323x?1x?1x?x?1x?13ABx?C令3，等式右边通分后比较两边分子x的同次项的系数得： ??2x?1x?1x?x?1解：??A?B?0?A?1??A?B?C?0 解此方程组得：??B??1?A?C?3?C??2??

31?x?21x?2 ????x3?1x?1x2?x?1x?1x2?x?1131313(2x?1)?(2x?1)(2x?1)x?22?2222而2 ?22???2222x?x?1x?x?1x?x?1x?x?1x?x?1x?x?133x11(2x?1)??3dx??22dx??dx??2dxx?1x?x?1x?12x?x?11x?11 22)?lnx?1?1?3?d(d(x?x?1)?x2?x?1123x?2)2?12(32??3arctan2x?11?lnx?1?ln(x2?x?1)?C

23 29

?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 30

百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读，免费范文网，提供经典小说综合文库中国人民大学出版社（第四版）高等数学一第4章课后习题详解(7)在线全文阅读。