Here's an old qual problem in analysis. Let $s=\Vert x \Vert$ and define $$f(x)=\frac{1}{s(1+s^{1/2}\log s)}$$, $x\in \mathbb{R}$. Find $1\le p\le \infty$ for which $f\in L^p(\mathbb{R}^2)$.
Attempt - It suffices to find $p$ for which $\int_{\mathbb{R}^2}\vert f\vert^p<\infty.$ $\int _{\mathbb{R}^2}\vert f^p\vert=\int _{\mathbb{R}^2}\frac{1}{\vert s(1+s^{1/2}\log s\vert^p )}\le \int \frac{1}{s^p}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{(x^2+y^2)^{p/2}}dxdy=\int_{-\infty }^{\infty}\int_{-\pi/2}^{\pi /2} y^{p-2}\sec^{2-p}(\theta)d\theta dy$, and that's the farthest I can do in this direction.
I would try Holder but I am not sure for which $p$ we have $\Vert 1/s \Vert_p<\infty$.
Follow your idea, $s = \|x\|$,
Consider $A = \{s < 2\}$, $B = \{s > 2\}$.
On $A$, we have $$f\le \dfrac{C}{s}$$ for some constant $C$.
because $$\inf_{0 \le s < 2} s^{1/2}\log s$$ is bounded.
$$\int_A |f|^p\le \int_{A} \dfrac{C}{s^p} dxdy = \int_{0}^{2\pi}\int_{0}^{2}\dfrac{C}{s^p}s ds d\theta = 2\pi\int_0^{2}\frac{C}{s^{p-1}}ds$$
it is convergent when $p < 2$.
On $B$, we have $s(1 + s^{1/2}\log s) \sim s^{3/2}\log s$ or in other word, there is a constant $C'$,
$$s(1 + s^{1/2}\log s)\ge C' ( s^{3/2}\log s)$$
when $s > 2$.
$$\int_B |f|^p \le C \int_B \dfrac{1}{(s^{3/2}\log s)^p} = 2\pi \int_2^{\infty} \dfrac{1}{s^{3p/2 - 1}(\log s)^p} ds$$
when $3p/2 - 1 > 1$, it is convergent for sure. i.e. $p > 4/3$.
if $p = 4/3$, you got
$$\int_2^{\infty}\frac{1}{s(\log s)^{4/3}}ds $$ which is convergent too.
thus
$$2 > p \ge 4/3.$$