How do I break down (x-2)/((x+1)(x-1)^2) into partial fractions?

Firstly, let (x-2)/((x+1)(x-1)²) = A/(x+1) + B/(x-1)+C/(x-1)², where A,B and C are to be calculated. 

Then, multiply both sides of the equation by (x+1)(x-1)² which will leave an equation as follows;

x-2 = A(x-1)² + B(x+1)(x-1) +C(x+1). Now let x=1.

-1 = 0A + 0B + 2C   this implies C = -1/2

Now let x = -1

-3 = A*(-2)²  + 0B + 0C  this implies -3=4*A which means A = -3/4

Now we have equation: x-2 = (-3/4)(x-1)²  + B(x-1)(x+1)  -(1/2)(x+1)

If we let x = 0 we get; 

-2 = -3/4 -B  - 1/2 giving; 

-2 = -5/4 - B          Adding 5/4 to both sides gives  -3/4 = - B from which we can see B = 3/4

now we have transformed our initial fraction into partial fractions; 

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

To check your answers enter in an arbitrary x, both sides should come to the same number.