Question

求不定积分：（x-5）dx/（x^3-3x^2+4)

Answer 1

x^3-3x^2+4=(x+1)(x^2-4x+4)=(x+1)(x-2)^2

用有理函数的方法

所以(x-5)/(x^3-3x^2+4)=A/(x+1)+B/(x-2)+C/(x-2)^2

两边同乘(x^3-3x^2+4)

得到

x-5=A(x-2)^2+B(x+1)(x-2)+C(x+1)

对于任意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

然后分开积分

∫(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