∫(x^3+1)dx/x^3-5x^2+6x

(x^3+1)/(x^3-5x^2+6x)
=[(x^3-5x^2+6x)+(5x^2-6x+1)]/(x^3-5x^2+6x)
=1+(5x^2-6x+1)/(x^3-5x^2+6x)
=1+(5x^2-6x+1)/[x(x-2)(x-3)].
用待定系数法：假设(5x^2-6x+1)/[x(x-2)(x-3)]=a/x+b/(x-2)+c/(x-3),对右边通分得
(5x^2-6x+1)=a(x-2)(x-3)+bx(x-3)+cx(x-2)=(a+b+c)x^2-(5a+3b+2c)x+6a,
比较各项系数知a+b+c=5,5a+3b+2c=6,6a=1,易解得a=1/6,b=-9/2,c=28/3,所以
(x^3+1)/(x^3-5x^2+6x)
=1+(5x^2-6x+1)/[x(x-2)(x-3)]
=1+(1/6)*1/x-(9/2)*1/(x-2)+(28/3)*1/(x-3).
两边积分得
∫ (x^3+1)/(x^3-5x^2+6x) dx
=x+(1/6)lnx-(9/2)ln(x-2)+(28/3)ln(x-3)+C,C为任意常数