Answer
$\frac{1}{4}\ln|8x-2|+c $
Work Step by Step
Let $ u=\ln |8x-2|$, then $ du= \frac{8}{8x-2} dx=\frac{4}{4x-1} dx $ and hence
$$\int \frac{1}{(4x-1)\ln (8x-2)}dx =\frac{1}{4} \int \frac{1}{u}du\\
=\frac{1}{4}\ln |u|+c
=\frac{1}{4}\ln|8x-2|+c.$$
