求(-x^2-2)/(x^2+x+1)^2dx的不定积分,用分部积分法

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黎哲妍多名
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是不是想求:∫{(2x^2+x+1)/[(x+3)(x-1)^2]}dx

 若是这样,则方法如下:
原式=∫{[(2x^2-4x+2)+(5x-1)]/[(x+3)(x-1)^2]}dx
  =2∫[1/(x+3)]dx+∫{[(5x-5)+4]/[(x+3)(x-1)^2]}dx
  =2∫[1/(x+3)]d(x+3)+5∫[1/(x+3)(x-1)]dx
   +4∫[1/(x+3)(x-1)^2]dx
  2∫[1/(x+3)]d(x+3)+5∫[1/(x+3)(x-1)]dx
   +4∫[1/(x+3)(x-1)^2]d(x-1)
  =2ln|x+3|+(5/4)∫{[(x+3)-(x-1)]/(x+3)(x-1)]}dx
   -4∫[1/(x+3)]d[1/(x-1)]
  =2ln|x+3|+(5/4)∫[1/(x-1)]dx-(5/4)∫[1/(x+3)]dx
   -4∫[1/(x+3)]d[1/(x-1)]
  =2ln|x+3|+(5/4)∫[1/(x-1)]d(x-1)-(5/4)∫[1/(x+3)]d(x+3)
   -4∫[1/(x+3)]d[1/(x-1)]
  =2ln|x+3|+(5/4)ln|x-1|-(5/4)ln|x+3|-4∫[1/(x+3)]d[1/(x-1)]
  =(3/4)ln|x+3|+(5/4)ln|x-1|-4∫[1/(x+3)]d[1/(x-1)]。
下面求∫[1/(x+3)]d[1/(x-1)]的值
令1/(x-1)=u,则:x-1=1/u,∴x=1+1/u=(u+1)/u,∴x+3=(4u+1)/u。
∴∫[1/(x+3)]d[1/(x-1)]
=∫[u/(4u+1)]du
=(1/4)∫[(4u+1-1)/(4u+1)]du
=(1/4)∫du-(1/4)∫[1/(4u+1)]du
=u/4-(1/16)∫[1/(4u+1)]d(4u+1)
=1/[4(x-1)]-(1/16)ln|4u+1|+c
=1/[4(x-1)]-(1/16)ln|4/(x-1)+1|+c
=1/[4(x-1)]-(1/16)ln|(x+3)/(x-1)|+c
=1/[4(x-1)]-(1/16)ln|x+3|+(1/16)ln|x-1|+c
∴原式=(3/4)ln|x+3|+(5/4)ln|x-1|
     -1/(x-1)+(1/4)ln|x+3|-(1/4)ln|x-1|+c
   =1/(1-x)+ln|x+3|+ln|x-1|+c
注:若原题不是我所猜测的那样,则请你补充说明。
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