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