Integration problem again

Let (x^2 + x - 3) / (x -1) (x - 2) (x + 1)=A/(x-1)+B/(x-2)+C/(x+1)
x^2 + x - 3=A(x-2)(x+1)+B(x-1)(x+1)+C(x-2)(x-1)
sub x=2
3B=3, B=1
sub x=-1
6C=-3,C=-1/2
sub x=1
-2A=-1
A=1/2
(x^2 + x - 3) / (x -1) (x - 2) (x + 1)=1/2(x-1)+1/(x-2)-1/2(x+1)
So
S ((2 (x^2 + x - 3)) / ((x -1) (x - 2) (x + 1)))dx
= S 1/(x-1)+2/(x-2)-1/(x+1) dx
=ln|x-1|+2ln|x-2|-ln|x+1|+C