i think this should be solvable by "by parts" but i am not getting an answer from that. Probably I am not doing it correctly.
thanks
This doesn't ask you to find the integral! ln(x) is an increasing function so for x between 2 and 3, $\displaystyle ln(2)\le ln(x)\le ln(3)$ and $\displaystyle 1/ln(3)\le 1/ln(x)\le 1/ln(2)$.
So $\displaystyle 1/ln(3)= \int_2^3 dt/ln(3)\le \int_2^3 \frac{dx}{ln(x)}\le 1/ln(2)= \int_2^3 dt/ln(2)$
Now, 1/ln(2)= 1.44 and 1/ln(3)= 0.9102 so that
$\displaystyle 0.9102\le \int_2^3 \frac{dx}{ln(x)}\le =1.44$