Let $$f(x) = \begin{cases} \frac{1}{|x|\log{(\frac{1}{|x|}})^2} & |x| \le 1/2 \\ 0 &\text{otherwise} \end{cases}$$
I want to show that $f(x)$ is Lebesgue integrable. First, by additivity of the integral, I get that
$$ \int_{\mathbb{R}} f(x)\, dx = \int_{[-\frac{1}{2},\frac{1}{2}]} \frac{1}{|x|\log{(\frac{1}{|x|}})^2 }\, dx$$
From here, I've tried getting different approximations, but I've failed. I'm not sure how to deal with the undefined value at $x=0$ and the asymptotes at $x=1$ and $x=-1$. Any ideas how to deal with these points?
Hint: That integral equals
$$2\int_0^{1/2}\frac{1}{x(\ln x)^2}\,dx.$$ Let $u=\ln x.$