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

We multiply throughout by `(x-1)^3(x+1)`:

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

We do not need to expand all this out. Instead, we just use appropriate substitutions to find the values of the unknowns `A` to `D`.

Let `x = 1`:

LHS `=-1`

RHS `= 2C`

So `C = -1/2`

Let `x = -1`:

LHS `= -1`

RHS `= -8D`

So `D = 1/8`

We now compare the coefficients of `x^3` on both sides and then compare the constant values on both sides.

Coefficient of `x^3` on LHS `= 0`

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

So `A=-D`.

But since `D = 1/8`, we have `A = -1/8`.

Constant term on LHS `= -3`

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

But since we know 3 values now, we have: `B = 9/4`.

So

`(2x^2-3)/((x-1)^3(x+1))` `=-1/(8(x-1))+9/(4(x-1)^2)` `-1/(2(x-1)^3)+1/(8(x+1)`

Easy to understand math videos:
MathTutorDVD.com