Weak convergence in $H^1(\mathbb R^3)$ implies convergence of integrals

Question

Suppose that $f_n \rightharpoonup f$ in $H^1(\mathbb R^3)$ (weak convergence). Then $$\int_{\mathbb R^3} \frac{\lvert f_n(x) \rvert^2}{\lvert x \rvert} dx \stackrel{n\to \infty}{\longrightarrow} \int_{\mathbb R^3} \frac{\lvert f(x) \rvert^2}{\lvert x \rvert}dx.$$ Approach: I tried to use the fact that $\mathbb 1_{\Omega} \,f_n(x) \to \mathbb 1_{\Omega} \, f_(x)$ in $L^2(\mathbb R^3)$ for each bounded $\Omega$: Note that $$ \left \lvert \int_{\mathbb R^3} \frac{\lvert f_n(x) \rvert^2}{\lvert x \rvert} dx - \int_{\mathbb R^3} \frac{\lvert f(x) \rvert^2}{\lvert x \rvert}dx\right \rvert = \left \lvert \int_{\mathbb R^3} \frac{\lvert f_n(x) \rvert^2 - \lvert f(x) \rvert^2}{\lvert x \rvert}dx\right \rvert \\ = \left \lvert \int_{|x| \leq R}\frac{\lvert f_n(x)|^2 - \lvert f(x)\rvert ^2}{|x|} dx + \int_{|x| > R}\frac{\lvert f_n(x)|^2 - \lvert f(x)\rvert ^2}{|x|} dx\right \rvert \\ \leq \int_{|x| \leq R} \frac{\lvert \lvert f_n(x)|^2 - \lvert f(x)\rvert ^2 \rvert }{|x|} dx + \frac{1}{R} \int_{|x| > R} \lvert \lvert f_n(x) \rvert - \lvert f(x) \rvert \rvert dx.$$ Now the left term can be bounded by the Sobolev inequality, but is that helpful? I suppose that not necessarily we have $\nabla f_n \mathbb 1_{\Omega} \to \nabla f \mathbb 1_{\Omega}$ in $L^2$. Any hints?