I am currently writing a report on Liouville's theorems on integration in finite terms, and I am in the process of proving the Liouville–Hardy theorem. This is what I understand so far.
Theorem Let $f(x)$ be a rational function then $\int f(x) \log(x) dx$ is elementary if and only if there exists a rational function $g(x)$ and a constant $c$ such that $$f(x)=\dfrac{c}{x}+g'(x).$$
Proof (attempt)
We note that this is another special case of the strong Liouville theorem with rational function $f(x)$, $y_{1}= \log(x)$ and $F(x,y_{1})=f(x)y_{1}$ F is clearly a rational function in $x,y_{1}$ and we observe that $\dfrac{dy_{1}}{dx}=\dfrac{1}{x}$ which is also a rational function. Applying the strong Liouville theorem we can say that for $\int f(x) \log(x) dx$ to be elementary we need
$$ \int f(x) \log(x) dx= u_{0}(x, \log(x))+ \sum\limits_{i=1}^k c_{i} \log(u_{i}(x, \log(x)), $$
where $u_{0}, u_{i}$ are rational functions in $x, \log(x)$ and $c_{i}$ is a constant for $1 \leq i \leq k$ differentiating both sideswith respect to x we get
$$ f(x)log(x)=\dfrac{d}{dx}[u_{0}(x, \log(x))+ \sum\limits_{i=1}^k c_{i} \log(u_{i}(x, \log(x))] $$
Next we consider $u_{0}(x, \log(x))$ and take a taylor expansion about $\log(x)=0$ this gives us:
$$\begin{align} u_{0}(x, \log(x)) & = u_{0}(x,0) + \dfrac{\partial u_{0}(x,0)}{\partial \log(x)} \log(x) + \dfrac{1}{2!}\dfrac{\partial^{2} u_{0}(x,0)}{ \partial \log(x)^{2}} \log^{2}(x) \\ & + \dots + \dfrac{1}{n!}\dfrac{\partial^{n} u_{0}(x,0)}{ \partial \log(x)^{n}} \log^{n}(x) + R, \end{align}$$
where $R$ is our remainder term, given by $R = \dfrac{1}{(n+1)!}\dfrac{\partial^{n+1} u_{0}(x,\zeta)}{( \partial \log(x)^{n+1}} \log^{n+1}(x)$ for some $1 < \zeta < \log(x)$
If you go here to page 303 http://www.rangevoting.org/MarchisottoZint.pdf the rest of the proof is here but I do not understand what is going on? Could anybody explain the rest to me please?