Answer
$$\int\frac{dx}{x(\log_{8}x)^2}=-\frac{(\ln8)^2}{\ln x}+C$$
Work Step by Step
$$A=\int\frac{dx}{x(\log_{8}x)^2}=\int\frac{dx}{\frac{x(\ln x)^2}{(\ln8)^2}}$$ $$A=(\ln8)^2\int\frac{dx}{x(\ln x)^2}$$
We set $u=\ln x$, which means $$du=\frac{dx}{x}$$
Therefore, $$A=(\ln8)^2\int\frac{1}{u^2}du$$ $$A=(\ln8)^2\times\Big(-\frac{1}{u}\Big)+C$$ $$A=-\frac{(\ln8)^2}{\ln x}+C$$
