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

We multiply both sides by `x^2(x-2)`:

`1 = Ax(x - 2) + B(x - 2) + Cx^2`

Let `x = 0`, and this gives `B = -1/2`.

Let `x = 2`, and this gives `C = 1/4`.

Comparing coefficients for `x^2`, we obtain `A + C = 0`.

So `A = -1/4`.

Therefore

`1/(x^2(x-2))`

`=A/x+B/x^2+C/(x-2)`

`=-1/(4x)-1/(2x^2)+1/(4(x-2))`

So the integral required is:

`int(dx)/(x^2(x-2))`

`=int(-1/(4x)-1/(2x^2)+1/(4(x-2)) )dx`

`=-1/4ln x+1/(2x)+1/4ln(x-2)+K`