(2x^2-3x-3)/[(x-1)(x^2-2x+5)]求不定积分

(2x^2-3x-3)/[(x-1)(x^2-2x+5)]求不定积分

设(2x²-3x-3)/[(x-1)(x²-2x+5)]=[a/(x-1)]+[(bx+c)/(x²-2x+5)]
则2x²-3x-3=a(x²-2x+5)+(bx+c)(x-1)
整理得
2x²-3x-3=(a+b)x²-(2a+b-c)x+(5a-c)
∴a+b=2
2a+b-c=3
5a-c=-3
解得
a=-1
b=3
c=-2
∴(2x²-3x-3)/[(x-1)(x²-2x+5)]=[-1/(x-1)]+[(3x-2)/(x²-2x+5)]

∴∫(2x²-3x-3)/[(x-1)(x²-2x+5)]dx
=∫[-1/(x-1)]dx+∫[(3x-2)/(x²-2x+5)]dx
=-ln|x-1|+∫[(3/2)(2x-2)/(x²-2x+5)]dx+∫[1/(x²-2x+5)]dx
=-ln|x-1|+(3/2)ln|x²-2x+5|+∫[1/((x-1)²+4)]dx
=-ln|x-1|+(3/2)ln|x²-2x+5|+(1/4)∫[1/(((x-1)/2)²+1)]dx
=-ln|x-1|+(3/2)ln|x²-2x+5|+(1/2)∫[1/(((x-1)/2)²+1)]d[(x-1)/2]
=-ln|x-1|+(3/2)ln|x²-2x+5|+(1/2)arctan[(x-1)/2]+C
C为任意常数

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