Properties of function $f(x) = (1 + x^2)^{-\alpha/2}(\log(2+x^2))^{-1},\text{ }x \in \mathbb{R}$ with $0 < \alpha < 1$.

Let $x>0$. Because $\log x\le x$, we have that $\log x\le x^t/t$ for $t>0$. Therefore $$\frac{1}{\log (2+x^2)}\ge \frac{t}{(2+x^2)^t},\ x\in\mathbb{R}.$$

Then $$\int_{\mathbb{R}} (1+x^2)^{-\alpha q/2}(\log (2+x^2))^{-q}\ge \int_{\mathbb{R}}(1+x^2)^{-\alpha q/2}(2+x^2)^{-qt}. $$

From the last inequality, you can conclude that if $q\in [1,1/\alpha)$ then $f\notin L^q(\mathbb{R})$.

On the other hand, note that $$\int_{\mathbb{R}} (1+x^2)^{-\alpha p/2}(\log (2+x^2))^{-p}\le C\int_{\mathbb{R}} (1+x^2)^{-\alpha q/2},\tag{1}$$

for some positive constant $C$ then, $f\in L^p(\mathbb{R})$ for $p>1/\alpha$. To conclude, note that $$f'(x)=-(2+\alpha)x(1+x^2)^{-1-\alpha/2}\left(\frac{1}{(\log(1+x^2))^2}+\frac{1}{\log(1+x^2)}\right).$$

As in $(1)$, there exist $D>0$ such that $$\int_{\mathbb{R}} |f'|^p\le D\int_{\mathbb{R}}|x|^p(1+x^2)^{-p-p\alpha/2},$$

whence $f'\in L^p(\mathbb{R})$ if $p> 1/(\alpha+1)$, which implies that $f\in W^{1,p}(\mathbb{R})$ for $p> 1/\alpha$.

The case $p=\infty$ is immediate.

Remark: The above calculation does not includes the case where $p=1/\alpha$, however, after thinking for a while, I noted that the change of variables $2+x^2=e^u$ prove all cases, in fact, after making this change of variables, you will have to estimate the following integral $$\int_{N}^\infty \frac{e^u}{(e^u-1)^{\alpha p/2}(e^u-2)^{1/2}u^{p}},$$

where $N$ is bigger thatn $\log 2$. Once $$\frac{e^u}{(e^u-1)^{\alpha p/2}(e^u-2)^{1/2}},$$

is bounded for $p\ge 1/\alpha$ and $1/u^{p}$ is integrable in $[N,\infty)$ for $\alpha\in (0,1)$, we have proved the desired result.