this can be written as`int_1^oo (lnx/x) dx`
this is type 1 improper integral which the bounds are between 1 and
using limit we can solve this:
`lim_(t->oo)int_1^tlnx/xdx`
let u=lnx, du=`1/xdx`
`lim_(t->oo)int_1^tudu` and integrate `lim_(t->oo)[u^2/2]_1^t`
In previous step we set u=ln x so substitute it into our equation and we have `lim_(t->oo)[(lnx)^2/2]_1^t` evaluate with t and 1
`lim_(t->oo)[ln(t)^2/2-ln(1)^2/2]`
`lim_(t->oo)[oo-0]=oo`
`:.` it is
and diverging.