Question

∫（4x-3）/（x-1）（x^2+2x+5）dx

求有理函数的积分

Answer 1

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

追问

这跟我的是同一道题吗？？？