Answer
$$\frac{{\ln 2{{\left( {{{\log }_2}\left( {5x + 1} \right)} \right)}^3}}}{{15}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {{{\log }_2}\left( {5x + 1} \right)} \right)}^2}}}{{5x + 1}}} dx \cr
& {\text{set }}u = {\log _2}\left( {5x + 1} \right){\text{ then }}\frac{{du}}{{dx}} = \frac{1}{{\ln 2}}\left( {\frac{5}{{5x + 1}}} \right) \to \frac{1}{{5x + 1}}dx = \frac{{\ln 2}}{5}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{{\left( {{{\log }_2}\left( {5x + 1} \right)} \right)}^2}}}{{5x + 1}}} dx = \int {{u^2}} \left( {\frac{{\ln 2}}{5}du} \right) \cr
& = \frac{{\ln 2}}{5}\int {{u^2}} du \cr
& {\text{using }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& = \frac{{\ln 2}}{5}\left( {\frac{{{u^3}}}{3}} \right) + C \cr
& = \frac{{\ln 2}}{{15}}{u^3} + C \cr
& {\text{replace }}{\log _2}\left( {5x + 1} \right){\text{ for }}u \cr
& = \frac{{\ln 2{{\left( {{{\log }_2}\left( {5x + 1} \right)} \right)}^3}}}{{15}} + C \cr} $$
