You can evaluate this integral by using integration by parts.
Let #u=ln(2x)#
Let #dv=1/x^2dx #
Differentiating #u=ln(2x)# we have
#du=1/xdx #
Integrating #dv=1/x^2dx# we have
#intdv=intx^-2dx#
#v=-1/x #
Recall the integration by parts formula
#uv-intvdu #
Now proceed as follows
#-ln(2x)/x-int-1/x(1/x)dx #
#-ln(2x)/x-int-1/x^2dx #
#-ln(2x)/x+intx^-2dx #
#-ln(2x)/x-1/x+C #
We can rewrite if you like
#-((ln(2x)+1))/x+C #
General note:
For #n!=-1#, any integral of the form #intx^nlnxdx# can be found by the process described above.
(For #n=-1# use #u#-substitution, with #u=lnx# so #du=1/xdx#.)
