Question

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

Answer 1

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`.)