How do you evaluate the integral of #ln(2x)/x^2 dx#?

1 Answer
Mar 16, 2015

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#.)