Find the minimum value of this integral: For what value of $k > 1$ is $$ \int_k^{k^2} \frac 1x \log\frac{x-1}{32}\, \mathrm dx $$ minimal?
After applying Newton-Leibniz, I got $k = 3$ and then did 2nd derivative test, it gave me positive result, thus 3 is the answer but I want to know if there's a smarter/slicker way to do?