Use partial fractions :

Let 1/(x-1)(x+1)(x+2) = A/(x-1) + B/(x+1) + C/(x+2)

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

For x = 1 : A = 1/6

For x = -1 : B = -1/2

For x = -2 : C = 1/3

so, 1/(x-1)(x+1)(x+2) = 1/6(x-1) - 1/2(x+1) + 1/3(x+2)

Integrating the above expression we get

(1/6)ln(x-1) - (1/2)ln(x+1) + (1/3)ln(x+2) + c

where c is the constant of integration