Answer
$$ - \frac{{{{\ln }^2}8}}{{\ln x}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x{{\left( {{{\log }_8}x} \right)}^2}}}} \cr
& {\text{or we can write}} \cr
& = \int {{{\left( {{{\log }_8}x} \right)}^{ - 2}}\left( {\frac{1}{x}} \right)} dx \cr
& {\text{using the property }}{\log _a}x = \frac{{\ln x}}{{\ln a}}{\text{ }}\left( {{\text{see example 7b}}} \right) \cr
& = \int {{{\left( {\frac{{\ln x}}{{\ln 8}}} \right)}^{ - 2}}\left( {\frac{1}{x}} \right)} dx \cr
& = \int {{{\left( {\frac{{\ln 8}}{{\ln x}}} \right)}^2}\left( {\frac{1}{x}} \right)} dx \cr
& = {\ln ^2}8\int {{{\left( {\frac{1}{{\ln x}}} \right)}^2}\left( {\frac{1}{x}} \right)} dx \cr
& {\text{use substitution}}{\text{. Let }}u = \ln x,{\text{ then }}du = \frac{1}{x}dx \cr
& {\text{write the integral in terms of }}u \cr
& = {\ln ^2}8\int {\frac{1}{{{u^2}}}} du \cr
& = {\ln ^2}8\int {{u^{ - 2}}} du \cr
& {\text{integrate using the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& = {\ln ^2}8\left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = - \frac{{{{\ln }^2}8}}{u} + C \cr
& {\text{write in terms of }}x{\text{; replace }}\ln x{\text{ for }}u \cr
& = - \frac{{{{\ln }^2}8}}{{\ln x}} + C \cr} $$
