x^3-3x^2+4=(x+1)(x^2-4x+4)=(x+1)(x-2)^2
用有理函數(shù)的方法
所以(x-5)/(x^3-3x^2+4)=A/(x+1)+B/(x-2)+C/(x-2)^2
兩邊同乘(x^3-3x^2+4)
得到
x-5=A(x-2)^2+B(x+1)(x-2)+C(x+1)
對(duì)于任意x都成立
令x=2
-3=3C,C=-1
令x=-1
-6=9A,A=-2/3
令x=0
-5=4A-2B+C
B=(5-8/3-1)/2=2/3
所以
(x-5)/(x^3-3x^2+4)=(-2/3)/(x+1)+(2/3)/(x-2)-1/(x-2)^2
然后分開積分
∫(x-5)/(x^3-3x^2+4)=-2/3 ln|x+1|+2/3 ln|x-2|+1/(x-2)+C
=(2/3)ln|(x-2)/(x+1)|+1/(x-2)+C