Answer
$$\frac{{9\ln 2}}{4}$$
Work Step by Step
$$\eqalign{
& \int_1^8 {\frac{{{{\log }_4}\theta }}{\theta }} d\theta \cr
& {\text{Use lo}}{{\text{g}}_a}x = \frac{{\ln x}}{{\ln a}}{\text{, so that }}{\log _4}\theta = \frac{{\ln \theta }}{{\ln 4}} \cr
& = \int_1^8 {\frac{{\ln \theta }}{{\ln 4}}\left( {\frac{1}{\theta }} \right)} d\theta \cr
& = \frac{1}{{\ln 4}}\int_1^8 {\ln \theta \left( {\frac{1}{\theta }} \right)} d\theta \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = \ln \theta,{\text{ so that }}du = \frac{1}{\theta }d\theta \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = 8,{\text{ then }}u = \ln 8 \cr
& \,\,\,\,\,\,{\text{If }}\theta = 1,{\text{ then }}u = 0 \cr
& {\text{Then}} \cr
& \frac{1}{{\ln 4}}\int_1^8 {\ln \theta \left( {\frac{1}{\theta }} \right)} d\theta = \frac{1}{{\ln 4}}\int_0^{\ln 8} u du \cr
& {\text{integrating}} \cr
& = \frac{1}{{\ln 4}}\left( {\frac{{{u^2}}}{2}} \right)_0^{\ln 8} \cr
& = \frac{1}{{2\ln 4}}\left( {{{\ln }^2}8 - 0} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{{{{\ln }^2}8}}{{\ln 16}} \cr
& = \frac{{{{\ln }^2}\left( {{2^3}} \right)}}{{\ln \left( {{2^4}} \right)}} \cr
& = \frac{{{3^2}{{\ln }^2}\left( 2 \right)}}{{4\ln \left( 2 \right)}} \cr
& = \frac{{9\ln 2}}{4} \cr} $$
