\frac{x^2-5x}{(x-1)(x+1)(x-2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}
Multiplying both sides by the common denominator:
x^2-5x=A(x+1)(x-2) + B(x-1)(x-2)+C(x-1)(x+1)
x^2-5x=A(x^2-x-2) + B(x^2-3x+2)+C(x^2-1)
x^2-5x=Ax^2-Ax-2A+Bx^2-3Bx+2B +Cx^2-C
(Ax^2+Bx^2+Cx^2-x^2)+(-Ax-3Bx+5x)+(-2A+2B-C)=0
(A+B+C-1)x^2+(-A-3B+5)x+(-2A+2B-C)=0
So now, you have 3 system of equations:
A+B+C = 1
-A-3B =-5
-2A+2B-C=0
Solving for A, B and C, you'll get:
A=2, B=1, C=-2
Substitute these back to the equation:
\frac{x^2-5x}{(x-1)(x+1)(x-2)}=\frac{2}{x-1}+\frac{1}{x+1}+\frac{-2}{x-2}
Now you can easily get the Integral of each fraction:
\int\frac{2}{x-1}+\int\frac{1}{x+1}+\int\frac{-2}{x-2}
=2ln|x-1|+ln|x+1|-2ln|x-2| +C
