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The problem is to find the value of $a>1$ at which this integral $$\int_{a}^{a^2} \frac{1}{x}\ln\Big(\frac{x-1}{32}\Big)dx$$

reaches its minimum value. I don't want to know the number, I just want feedback on the ideas I'm trying. Considering that $$f(x)=\frac{1}{x}\ln\Big(\frac{x-1}{32}\Big)<0$$

whenever $x<33$ and the fact that $f(x)=0$ at $x_0=33$, we have that all the area associated with the graph of this function is negative until the point $x_0$. So we want to find the value of $a$ which yields the greatest portion of this "negative area". The fundamental theorem of calculus states that $$\int_a^bf(x) = F(b) - F(a)$$

which gives us an expression which can be differentiated. Particularly, the first derivative is $$f' =\frac{1}{a^2}\ln\Big(\frac{a^2-1}{32}\Big)-\frac{1}{a}\ln\Big(\frac{a-1}{32}\Big)$$

or $$f'=\frac{1}{a^2}\ln\Big(\frac{a^2-1}{32}\Big)\cdot 2a-\frac{1}{a}\ln\Big(\frac{a-1}{32}\Big)\cdot 1$$

I think the second is correct since, per the fundamental theorem, we are applying the primitive $F$ to the upper and lower bound functions $a^2$, and $a$ - so this is a sum of two composite functions differentiated by the chain rule. It depends on the correct interpretation of $a$ I think.

If we differentiate again, we can find the values of $a$ for which $f(a)''>0$. This will be value of $a$ where the integral reaches its minimum value. However, applying the second derivative test to an integral doesn't seem proper since it is a tool for studying the concavity of $f$ - what would be the interpretation here?

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Your second expression for $f'(a)$ is correct. Also, you shouldn't differentiate it again, but rather solve for the critical points of $f'(a)$ to find the minimum value of the integral because the integral is $f(a)$: $$f'(a)=0=\frac{1}{a}\left(2\ln{\left(a^2-1\right)}-2\ln{32}-\ln{\left(a-1\right)}+\ln{32}\right)$$ $$ \ln{32}=\ln{\left(a-1\right)}+2\ln{\left(a+1\right)}$$

Now, solve for $a$ then test if its a minimum or maximum by using the first derivative test.

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