Answer
\[ = \frac{{\,{{\left( {\ln 2x} \right)}^2}}}{2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{\ln 2x}}{x}\,dx} \hfill \\
\hfill \\
set\,\,\,\,\ln \,2x = \,u\,\,\,and\,\,\,\,\frac{1}{{2x}}2\,dx = du\,\,or\,\frac{{dx}}{x} = du \hfill \\
\hfill \\
\operatorname{int} egral\,becomes \hfill \\
\hfill \\
\int_{}^{} {\frac{{\ln 2x}}{x}\,dx} = \int_{}^{} {u\,du} \hfill \\
\hfill \\
{\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= \frac{{{u^2}}}{2} + C \hfill \\
\hfill \\
substituting\,back\,\,\,u = \ln 2x \hfill \\
\hfill \\
= \frac{{\,{{\left( {\ln 2x} \right)}^2}}}{2} + C \hfill \\
\end{gathered} \]