Improper integral $\int_{\frac{3}{2}}^\infty\frac{\log{(1+\frac{1}{x}})}{|x-2|^{\alpha+2}}\, dx$, problem with absolute value

Question

For discussing the convergence of the integral $$\int_{\frac{3}{2}}^\infty\frac{\log{(1+\frac{1}{x}})}{|x-2|^{\alpha+2}}\, dx$$ I have tried to do the following:

1. I want to observe that the "problem" of this integral, and this is the reason why it is an improper integral, is in the fact that the integration interval is no-bounded, it is $[\frac{3}{2},+\infty]$ and the fact that $\displaystyle\lim_{x\to 2}f(x)=+\infty$.
2. I want to eliminate the absolute value and I have tried to split the integral in this way: $$\int_{\frac{3}{2}}^\infty\frac{\log{(1+\frac{1}{x}})}{|x-2|^{\alpha+2}}\, dx=\int_{\frac{3}{2}}^2\frac{\log{(1+\frac{1}{x}})}{(2-x)^{\alpha+2}}\, dx+\int_{2}^3\frac{\log{(1+\frac{1}{x}})}{(x-2)^{\alpha+2}}\, dx+\int_{3}^\infty\frac{\log{(1+\frac{1}{x}})}{(x-2)^{\alpha+2}}\, dx $$ As matter of fact $(x-2)\geq 0\iff x\geq 2$ and so $|x-2|=\begin{cases} x-2 \, ,x\geq 2\\ 2-x \, , x\leq 2\end{cases}$.

$\textbf{My question:}$ it is right what I have done or the first two integral of the splitting must be of the following kind? $$\int_{\frac{3}{2}}^{2-\epsilon}\frac{\log{(1+\frac{1}{x}})}{(2-x)^{\alpha+2}}\, dx+\int_{2-\epsilon}^3\frac{\log{(1+\frac{1}{x}})}{(x-2)^{\alpha+2}}\, dx$$
In effect I have not understood a thing: when I consider $\int_{\frac{3}{2}}^2\frac{\log{(1+\frac{1}{x}})}{|x-2|^{\alpha+2}}\, dx$, then it means that $x\in[\frac{3}{2},2]$ or $x\in (\frac{3}{2},2)$ and so can I write $\displaystyle\int_{\frac{3}{2}}^2\frac{\log{(1+\frac{1}{x}})}{|x-2|^{\alpha+2}}\, dx=\int_{\frac{3}{2}}^2\frac{\log{(1+\frac{1}{x}})}{(2-x)^{\alpha+2}}\, dx$ ?

$\textbf{EDIT:}$ maybe my question is not clear. I would like to know if what I have written in 2) is right...then the idea is to consider in the first integral the behaviour of the function when $x\to 2^-$ and for the second $x\to 2^+$. For the reason why that in both cases I have unbounded functions respectively of this kind: $$\frac{\log{(1+\frac{1}{x})}}{-(x-2)^{\alpha+2}}\sim \frac{\log{1+\frac{1}{2^{- }}}}{-(x-2)^{\alpha+2}}\text{ as}\, x\to 2^{-}$$ and $$\frac{\log{(1+\frac{1}{x})}}{(x-2)^{\alpha+2}}\sim \frac{\log{1+\frac{1}{2^{+ }}}}{(x-2)^{\alpha+2}}\text{ as}\, x\to 2^{+}$$ I observe that the exponent of the terms $(x-2)$ and $(2-x)$ is $\alpha+2$ so the convergence of these two integrals is satisfied for $\alpha+2<1$. Then it remains to treat $\int_{3}^\infty\frac{\log{(1+\frac{1}{x}})}{(x-2)^{\alpha+2}}\, dx$