Prove that $u(x)=\log\log\left(1+\frac{1}{\vert x\vert}\right)\in H^1(B(0,1))$

Problem: Let $$\Omega\subset\mathbb R^2$$ denote the open unit ball in $$\mathbb R^2$$. Prove that the unbounded function $$f(x)=\log\log\left(1+\frac{1}{\vert x\vert}\right)$$ belongs to $$H^1(\Omega).$$

My Attempt: Let $$\{\varepsilon_n\}_{n=1}^\infty\subset[0,1]$$ such that $$\varepsilon_n\searrow0$$ as $$n\to\infty$$. Put $$\Omega_n=B(0,\varepsilon_n)$$. Define the sequence of functions $$f_n(x)=\begin{cases}f(x)&\text{if }x\in\Omega\setminus\Omega_n\\0&\text{otherwise.}\end{cases}$$ Note that $$\vert f_n\vert^2\nearrow\vert f\vert^2$$ as $$n\to\infty$$, so by the monotone convergence theorem we have $$\|f_n\|_{L^2(\Omega)}^2\to\|f\|_{L^2(\Omega)}^2$$ as $$n\to\infty$$. Using integration in polar coordinates, as shown in Folland's Real Analysis text, we have that \begin{align*} \|f_n\|_{L^2(\Omega)}^2 &=\int_{\Omega\setminus\Omega_n} \vert f(x)\vert^2\,dx=\int_{\Omega\setminus\Omega_n}\left\vert\log\log\left(1+\frac{1}{\vert x\vert}\right)\right\vert^2\,dx\\ &=2\pi\int_{\varepsilon_n}^1 r\left\vert\log\log\left(1+\frac1r\right)\right\vert^2\,dr\\ &\leq2\pi\int_{\varepsilon_n}^1 e^r\,dr\\ &\leq2\pi\int_0^1e^r\,dr\\ &=2\pi e\\ &<\infty. \end{align*} Since the bound above does not depend on $$n$$, letting $$n\to\infty$$ shows that $$f\in L^2(\Omega)$$, by the monotone convergence theorem.
Next, observe that $$\nabla f(x)=\left(-\frac{x_1}{\log\left(1+\frac{1}{\vert x\vert}\right)(1+\vert x\vert)\vert x\vert^2},-\frac{x_2}{\log\left(1+\frac{1}{\vert x\vert}\right)(1+\vert x\vert)\vert x\vert^2}\right),$$ so that $$\vert\nabla f(x)\vert^2=\frac{1}{\log\left(1+\frac1{\vert x\vert}\right)^2(1+\vert x\vert)^2\vert x\vert^2}.$$ Using the same method as above we have that $$\|\nabla f_n\|_{L^2(\Omega)}^2\to\|\nabla f\|_{L^2(\Omega)}^2$$ by the monotone convergence theorem. Then, integrating in polar coordinates once again, we have \begin{align*} \|\nabla f_n\|_{L^2(\Omega)}^2 &=\int_{\Omega\setminus\Omega_n}\vert\nabla f(x)\vert^2\,dx=\int_{\Omega\setminus\Omega_n}\frac{1}{\log\left(1+\frac1{\vert x\vert}\right)^2(1+\vert x\vert)^2\vert x\vert^2}\,dx\\ &=2\pi\int_{\varepsilon_n}^1\frac{1}{\log\left(1+\frac1{r}\right)^2(1+r)^2r^2}\,dr\\ &\to\infty\quad\text{as }n\to\infty. \end{align*} It follows that $$f\notin H^1(\Omega)$$ since $$\vert\nabla f\vert\notin L^2(\Omega)$$.

Do you agree with my proof above? I am not sure that I fully understood and applied the definition of the Sobolev Space $$H^1(\Omega)$$, especially in the second part of the proof. Any clarification if I am in the wrong would be much appreciated.
Thank you for your time and valuable feedback.

In the computation of $$\int_{\Omega}|\nabla f|^2 \,dx$$ you forgot $$r\,dr$$.
Hence the integral converges since $$\frac{1}{\log(1+1/r)^2 r(1+r)^2}$$ is integrable close to $$r=0$$ and we get $$f\in H^1(\Omega)$$.