Answer
$$4$$
Work Step by Step
$$\eqalign{
& \int_1^e {\frac{{8\ln 3{{\log }_3}\theta }}{\theta }} d\theta \cr
& {\text{Use lo}}{{\text{g}}_a}x = \frac{{\ln x}}{{\ln a}}{\text{, so that }}{\log _3}\theta = \frac{{\ln \theta }}{{\ln 3}} \cr
& = 8\ln 3\int_1^e {\left( {\frac{{\ln \theta }}{{\ln 3}}} \right)\frac{1}{\theta }} d\theta \cr
& = 8\int_1^e {\left( {\ln \theta } \right)\frac{1}{\theta }} d\theta \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = \ln \theta,{\text{ so that }}du = \frac{1}{\theta }d\theta \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = e,{\text{ then }}u = \ln e = 1 \cr
& \,\,\,\,\,\,{\text{If }}\theta = 1,{\text{ then }}u = \ln 1 = 0 \cr
& {\text{Then}} \cr
& 8\int_1^e {\left( {\ln \theta } \right)\frac{1}{\theta }} d\theta = 8\int_0^1 u du \cr
& {\text{integrating}} \cr
& = 8\left( {\frac{{{u^2}}}{2}} \right)_0^1 \cr
& = 4\left( {{1^2} - {0^2}} \right) \cr
& {\text{simplifying, we get:}} \cr
& = 4 \cr} $$
