Answer
$\frac{2(2x- 1)}{(x-1)}$
Work Step by Step
$\frac{4x}{x-1} - \frac{2}{x+1} - \frac{4}{x^2 - 1}$
Since $x^2 - 1 = (x+1)(x-1)$
$=\frac{4x}{x-1} - \frac{2}{x+1} - \frac{4}{(x+1)(x-1)}$
$=\frac{4x \times (x+1)}{(x-1) \times (x+1)} - \frac{2 \times (x-1)}{(x+1) \times (x-1)} -
\frac{4}{(x+1)(x-1)}$
$=\frac{4x^2 + 4x -2x +2- 4}{(x+1)(x-1)}$
$=\frac{4x^2 +2x -2}{(x+1)(x-1)}$
$=\frac{2\times (2x^2 +x -1)}{(x+1)(x-1)}$
$=\frac{2\times (2x^2 +(2-1)\times x -1)}{(x+1)(x-1)}$
$=\frac{2(2x^2 +2x - x -1)}{(x+1)(x-1)}$
$=\frac{2(2x(x + 1) -1( x +1))}{(x+1)(x-1)}$
$=\frac{2((2x- 1)( x +1))}{(x+1)(x-1)}$
$=\frac{2(2x- 1)}{(x-1)}$
