With some simple rearrangement, we can rewrite the numerator as
[tex]2x^3 - 3x^2 - x + 4 = 2(x^3 - x) - 3x^2 + x + 4 \\\\ ~~~~~~~~ = 2x(x^2-1) - 3(x^2 - 1) + x + 1 \\\\ ~~~~~~~~ = (2x-3)(x^2-1) + x+1[/tex]
Then factorizing the difference of squares, [tex]x^2-1=(x-1)(x+1)[/tex], we end up with
[tex]\dfrac{2x^3 - 3x^2 - x + 4}{x^2 - 1} = \dfrac{(2x-3)(x-1)(x+1) + x+1}{(x-1)(x+1)} \\\\ ~~~~~~~~ = \boxed{2x-3 + \dfrac1{x-1}}[/tex]
