$\displaystyle

\int \frac{2x-1}{(4x^2-4x)^{2}} dx\

$

In this case what would u be? that the integral would be

$\displaystyle

\frac{1}{16} \, \int u^{-2} du

$

At first glance i thought it would just be

$\displaystyle

u=4x^2-4x

$

but obviously thats not it.