Answer
$ \displaystyle \frac{4x+1}{(2x-1)(x+1)}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$
$u(x)=(-2x+1)(x+1)\qquad $... , a product,
$... (fg)^{\prime}=f^{\prime}g+fg^{\prime} $
$\displaystyle \frac{du}{dx}=(-2)(x+1)$+$(-2x+1)(1)$
$=-2x-2-2x+1=-4x-1$
$\displaystyle \frac{d}{dx}[\ln|(-2x+1)(x+1)|]=$
$=\displaystyle \frac{1}{(-2x+1)(x+1)}\cdot(-4x-1)$
$=\displaystyle \frac{-4x-1}{(-2x+1)(x+1)}=\frac{-(4x+1)}{-(2x-1)(x+1)}$
$=\displaystyle \frac{4x+1}{(2x-1)(x+1)}$
