Find the minimum value of this integral.

Let $I:(1, \infty) \to \mathbb{R}$ be defined by $$ I(k)=\int_k^{k^2}\frac{1}{x}\log\frac{x-1}{32}dx. $$ Then $$ I'(k)=\frac{2}{k}\log\frac{k^2-1}{32}-\frac{1}{k}\log\frac{k-1}{32}=\frac{1}{k}\log\frac{(k+1)^2(k-1)}{32}. $$ We have $I'(k)<0=I'(3)$ for $1<k<3$, $I'(k)>0=I'(3)$ for $k>3$, therefore $I(3)$ is the minimal value of $I(k)$ for $k>1$.

$J=\int_k^{k^2}dx \frac{1}{x}\ln\frac{x-1}{32}=-\operatorname{dilog}(k)-\ln(\frac{k}{32}-\frac{1}{32})\ln(k)+\operatorname{dilog}(k^2)+\ln(\frac{k^2}{32}-\frac{1}{32})\ln(k^2)$ 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 $k=3$