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.)