How do you evaluate the integral of ln(2x)x2dx?

1 Answer
Mar 16, 2015

You can evaluate this integral by using integration by parts.

Let u=ln(2x)

Let dv=1x2dx

Differentiating u=ln(2x) we have

du=1xdx

Integrating dv=1x2dx we have

dv=x2dx

v=1x

Recall the integration by parts formula

uvvdu

Now proceed as follows

ln(2x)x1x(1x)dx

ln(2x)x1x2dx

ln(2x)x+x2dx

ln(2x)x1x+C

We can rewrite if you like

(ln(2x)+1)x+C

General note:
For n1, any integral of the form xnlnxdx can be found by the process described above.

(For n=1 use u-substitution, with u=lnx so du=1xdx.)