TITLE: Find the minimum value of this integral.
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: $J=\int_k^{k^2}dx \frac{1}{x}\ln\frac{x-1}{32}=-\operatorname{dilog}(k)-\ln(\frac{k}{32}-\frac{1}{32})\ln(k)+\operatorname{dilog}(k^2)+\ln(\frac{k^2}{32}-\frac{1}{32})\ln(k^2)$
where:
$$\operatorname{dilog}(x)=\int_1^x \, dt \ln(\frac{t}{1-t})$$
The derivative respect to $k$ is:
$$\frac{\partial J}{\partial k}=-\frac{1}{k}[5 \ln(2)+\ln(k-1)-2\ln((k-1)(k+1))$$
putting:
$$\frac{\partial J}{\partial k}=0$$
the result is $k=3$