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-用组合积分法求解四类有理函数的积分
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求定积分的方法
求定积分的方法
要 ·····································································································································3 关键词 ·····································································································································3 Abstract ·································································································································3 Keywords ······························································································································3 前言 ··········································································································································3
1. 定义法求定积分 ·········································································································3
定义法 ····················································································································3
典型例题 ················································································································4
2. 换元法求定积分 ·········································································································5
换元积分法 ············································································································5
典型例题 ················································································································5
3. 分部法求定积分 ·········································································································8
分部积分法··································································································· 8
典型例题······································································································· 8
4. 区间性质求定积分 ····································································································9
常见的三种题型··························································································· 9
典型例题······································································································· 9
5. 有理函数求积分 ······································································································ 11
有理函数积分法························································································· 11
典型例题····································································································· 11 参考文献 ······························································································································ 13
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PPT制作技巧导读：§3有理函数和可化为有理函数的不定积分，教学目的：掌握有理函数、三角函数及简单无理函数化有理函数积分的方法，重点难点：重点与难点为有理函数的分解，至此我们已经学得了一些最基本的积分方法．在此基础上，本节将讨论某些特殊类型的不定积分，这些不定积分无论怎样复杂，原则上都可按一定的步骤把它求出来．一有理函数的不定积分，有理函数是指由两个多项式函数的商所表示的函数，假分式总能化为一个多项式与一个真分式§3
有理函数和可化为有理函数的不定积分 教学目的：掌握有理函数、三角函数及简单无理函数化有理函数积分的方法。 重点难点：重点与难点为有理函数的分解。 教学方法：讲练结合。
至此我们已经学得了一些最基本的积分方法．在此基础上，本节将讨论某些特殊类型的不定积分，这些不定积分无论怎样复杂，原则上都可按一定的步骤把它求出来．
有理函数的不定积分 有理函数是指由两个多项式函数的商所表示的函数，其一般形式为 R(x)?P(x)Q(x)??0x??1x?0xmnn?1m?1????n????m??1x，
(1) 其中n，m为非负整数，?0,?1,?,?n与?0,?1,??m都是常数，且?0?0，?0?0． 若m?n，则称它为真分式；若m?n，则称它为假分式．由多项式的除法可知，假分式总能化为一个多项式与一个真分式之和．由于多项式的不定积分是容易求得的，因此只需研究真分式的不定积分，故设(1)为一有理真分式．
根据代数知识，有理真分式必定可以表示成若干个部分分式之和(称为部分分式分解)．因而问题归结为求那些部分分式的不定积分．为此，先把怎样分解部分分式的步骤简述如下(可与例1对照着做)：
对分母Q?x?在实系数内作标准分解：
Q?x???x?a1???x?as?其中?0?1,?i,?stij?1?2?x2?p1x?q1??1??x2?pt?qt??t，（2） ?i?1,2,?,t?均为自然数，而且
??i?1?2??j?1j?m;pj?4qj?0,j?1,2,?,t. 2
根据分母的各个因式分别写出与之相应的部分分式：对于每个形如?x?a?的因式，它所k对应的部分分式是
A1x?a?A2???Ak; ?x?a?k2?x?a?k对每个形如?x2?px?q?的因式，它所对应的部分分式是
B1x?C1x?px?q2?B2x?C2?x2?px?q?2???Bkx?Ck?x2?px?q?k. 把所有部分分式加起来，使之等于R?x?．(至此，部分分式中的常数系数Ai,Bi,Ci尚为待定的.)
确定待定系数：一般方法是将所有部分分式通分相加，所得分式的分母即为原分母Q?x?，第八章第三节第1页
而其分子亦应与原分子P?x?恒等．于是，按同幂项系数必定相等，得到一组关于待定系数的线性方程，这组方程的解就是需要确定的系数． 例1
对R?x??2x?x?4x?9x?10x?x?5x?2x?4x?作部分分式分解 解
按上述步骤依次执行如下：
Q?x??x5?x4?5x3?2x2?4x?8
??x?2??x?2??x2?x?1.? 2部分分式分解的待定形式为
R?x??A0x?2?A1x?2?A2?Bx?Cx?x?12?x?2?2.
（3） 用Q?x?乘上式两边，得一恒等式
2x?x?4x?9x?10?A0?x?2??x?x?1? 43222
+A1?x?2??x?2??x2?x?1??A2?x?2??x2?x?1?
+?Bx?C??x?2??x?2?
2然后使等式两边同幂项系数相等，得到线性方程组： ?A0?A1?B?2,????????x4的系数?33A?A?A?2B?C??1，???x的系数012??2???x的系数 ?A0?3A1?3A2?4B?2C?4，?4A1?3A2?8B?4C?9，?????x的系数???4A0?4A1?2A2?8C??10.????常数项求出它的解：A0?1,A1?2,A2??1,B??1,C?1，并代人(3)式，这便完成了R(x)的部分分式分解： R(x)?1x?2?2x?2?1(x?2)2?x?1x?x?12.
上述待定系数法有时可用较简便的方法去替代．例如可将x的某些特定值(如Q(x)?0的根)代人(4)式，以便得到一组较简单的方程，或直接求得某几个待定系数的值．对于上例，若分别用x?2和x??2代人(4)式，立即求得 A0?1和A2??1 于是（4）式简化成为 x?3x?12x?16?A1(x?2)(x?2)(x?x?1) 第八章第三节第2页
?(Bx?C)(x?2)(x?2)2. 为继续求得A1,B,C，还可用x的三个简单值代人上式，如令x?0,1,?1，相应得到 ?A1?2C?4,??A1?3B?3C?2, ?3A?B?C?8.1?由此易得A1?2,B??1,C?1．这就同样确定了所有待定系数．
一旦完成了部分分式分解，最后求各个部分分式的不定积分．由以上讨论知道，任何有理真分式的不定积分都将归为求以下两种形式的不定积分： (?)?(x?a)dxk；
?????Lx?M(x?px?q)2kdx(p?4q?0)． 2对于???，已知 ?lnx?a?C,?1???C,??1?k??x?a?k?1?p2?(x?a)dxk?1,k?1.k 对于????，只要作适当换元(令t?x? )，便化为 ?(xLx?M2?px?q)kdx??Lt?N?t2?r2?kdt ?L?2t(t?r)22kdt?N?dt(t?r)22k,
（5） 其中r2?q?p4,N?M?p2L.． 当k?1时，（5）式右边两个不定积分分别为
?t?rdt22dt??121ln(t?r)?C， 22?r2tarctan?C.
（6） rr当k?2时，（5）式右边第一个不定积分为
?(tt2?r)2kdt?12(1?k)(t?r)22k?1?C. 对于第二个不定积分，记
Ik? ?(tdt2?r)2k?1, 第八章第三节第3页 可用分部积分法导出递推公式如下：
Ik?1r1r22?(t?r)?t(t?r)1r222k222dt
?Ik?1??(t1t222k?r)dt
?1r2Ik?1??1?td2??(t2?r2)k?12r(k?1)?1?? ??
?经整理得到
Ik?1r2Ik?1???t?I. k?1??222k?12r(k?1)?(t?r)?t2r(k?1)(t?r)222k?1?2k?32r(k?1)2Ik?1.
（7） 重复使用递推公式(7)，最终归为计算I1，这已由(6)式给出.
把所有这些局部结果代回(5)式，并令t?x?x?122p2，II）的计算． 例2 求?(x2?2x?2)2dx. 解 在本题中，由于被积函数的分母只有单一因式，因此，部分分式分解能被简化为
x?1(x?2x?2)22?(x?2x?2)?(2x?1)(x?2x?2)222 ?1x?2x?22?2x?1(x?2x?2)22. 现分别计算部分分式的不定积分如下：
?xdx2?2x?22x?1??(x?1)d(x?1)2?1?arctanx(?1)?C1.
?(x2?2x?2)dx?2?(x(2x?2)?12?2x?2)22dx
??(x2d(x?2x?2)2?2x?2)2??d(x?1)?(x?1)22?1?2
??1x?2x?2??(tdt2?1). 第八章第三节第4页
由递推公式（7）,求得其中
?(tdt2?1)2?t2(t?1)2?1?2tdt2?112
x?12(x?2x?2)2?arctanx(?1)?C2. ?(xx?122?2x?2)dx?2x?32(x?2x?2)2?32arctanx(?1)?C.
下面再介绍几类被积函数能变换为有理数的不定积分。
二 三角函数有理式的不定积分 由u(x)、v(x)及常数经过有限次四则运算所得到的函数称为关于u(x)、v(x)的有理式，并用R(u(x),v(x))表示。 ?R(sinx,cosx)dx是三角函数有理式的不定积分。一般通过变换t?tanx2，可把它化为有理函数的不定积分。这是因为 xxx2sincos2tan22?2?2t,
sinx?21?t2x2x2xsin?cos1?tan222cos2x2x2?sin2x2?x21?tan2x2?1?t,
cosx?sin2?cos21?tan2
(10) 2?2t1?t所以?R(sinx,cosx)dx??R??1?t2,1?t2??2??1?t2dt． ?例3 求?1?sinxsinx(1?cosx)x2dx 解
令t?tan，将(8)、(9)、(10)代人被积表达式， 第八章第三节第5页
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有理函数积分的一些技巧
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