Firstly, we need to factor the denominator. We just use difference of 2 squares, twice:

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

So `(x^3-2)/(x^4-1)=(x^3-2)/((x^2+1)(x+1)(x-1)`

The partial fraction decomposition will be of the form:

`(x^3-2)/((x^2+1)(x+1)(x-1))` `=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1`

We multiply throughout by `(x^2 + 1)(x + 1)(x - 1)`:

`x^3 - 2 ` `-=(Ax + B)(x + 1)(x - 1)` ` + C(x^2 + 1)(x - 1)` ` + D(x^2 + 1)(x + 1)`

Let `x = 1`:

LHS `= -1`

RHS `= 4D`

So `D = -1/4`

Let `x = -1`:

LHS `= -3`

RHS `= -4C`

So `C = 3/4`

Coefficient of `x^3` on LHS `= 1`

Coefficient of `x^3` on RHS `= A + C + D`

Since `D = -1/4` and `C = 3/4`, then `A = 1/2`.

Constant term on LHS `= -2`

Constant term on RHS `= -B - C + D`

This gives us `B = 1`.

So

`(x^3-2)/((x^2+1)(x+1)(x-1))`

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

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