I am trying to "solve" this integral $$ \int_a^b \left( \frac{f'(x)}{f(x)} \right)^2 dx $$ where $0<a<b$, $f$ is continuous and diffentiable, $f'(x)>0$ and $f(a)>0$.
If the exponent would change from 2 to 1, the integral would be $\ln f(b)-\ln f(a)$. But it is that very term the source of my difficulty.
Your question is equivalent to determine $I(g) = \int_a^b (g'(x))^2 dx$ for any differentiable $g$.
1) Assume we can determine $I(g)$ for any for any differentiable $g$.
We know that $f(a) > 0$ and $f'(x) > 0$. Hence $f$ is strictly monotonically increasing, in particular $f(x) > 0$ on $[a,b]$. Therefore $g(x) = \ln(f(x))$ is well-defined and $g'(x) = \frac{f'(x)}{f(x)}$.
2) Assume we can determine your integral for any $f$ as above.
Let $g$ be differentiable. Define $f(x) = e^{g(x)}$. Then $f'(x) = g'(x)e^{g(x)}$, hence $\frac{f'(x)}{f(x)} = g'(x)$.
Now we have the problem that the derivative $g'(x)$ is not necessarily integrable. See What is an example that a function is differentiable but derivative is not Riemann integrable. In other words, we must impose suitable conditions on $g$ resp. $g'$.
Let us focus on continuously differentiable $g$. Determining $I(g)$ for any continuously differentiable $g$ is equivalent to determine $J(h) = \int_a^b h(x)^2 dx$ for any continuous $h$. But there is no solution in terms of $h$ or $\int_a^b h(x) dx$. See how to calculate integral of square of a function.
