I cannot get this identity by using the condition:
$\frac{1}{x^{2}}=-\partial(\frac{1}{x})$, and the integrands are defined by continuity at $x=0$.
My reasoning goes as $$\langle -\partial\frac{1}{x},\phi\rangle=\langle \frac{1}{x},\partial \phi\rangle=\int^{\infty}_{-\infty}\frac{1}{x^{2}}\phi dx$$ and we can break it into $$\lim_{\epsilon\rightarrow 0}\int^{\infty}_{\epsilon}\frac{1}{x^{2}}\phi dx+\int^{-\epsilon}_{-\infty}\frac{1}{x^{2}}\phi dx+\int^{\epsilon}_{-\epsilon}\frac{1}{x^{2}}\phi dx$$ The first and second term adds up to $$\int^{\infty}_{\epsilon} \frac{\phi(x)+\phi(-x)}{x^{2}}dx$$ while the third term $$\int^{\epsilon}_{-\epsilon}\frac{1}{x^{2}}\phi(x) dx=\int^{\epsilon}_{0}\frac{\phi(x)+\phi(-x)}{x^{2}}$$ approximates $$2\int^{\epsilon}_{0}\frac{\phi(0)}{x^{2}}$$
But this last step is not really justified; and even if it works the result looks remarkably different from the one I wanted. So I decided to ask.