This means that our integral becomes $$ I=\int_0^{\infty} r_1^2dr_1 \int_0^{\infty}r_2^2dr_2 \int_0^{\pi}dcos(\theta_1)\int_0^{\pi}dcos(\theta_2)\int_0^{2\pi}d\phi_1\int_0^{2\pi}d\phi_2 \frac{\exp{-2\alpha (r_1+r_2)}}{r_{12}} $$ where we have defined $$ \frac{1}{r_{12}}= \frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\beta)}} $$ with $$ \cos(\beta) = \cos(\theta_1)\cos(\theta_2)+\sin(\theta_1)\sin(\theta_2)\cos(\phi_1-\phi_2)) $$