How can I go from $$\rho(r_1)=2A^2\exp(-4\beta r_1^2)\int d\mathbf r_2 \exp(-4\beta r_2^2)\exp[(r_1^2+r_2^2-2r_1r_2\cos\theta)^{1/2}] \tag{1}$$ to $$\rho(r_1)=(4\pi A^2/r_1)\exp(-4\beta r_1^2+r_1)\int_0^{\infty}dr_2 r_2\,\exp(-4\beta r_2^2+r_2)\{ (r_1+r_2-1)+(1-\lvert r_1-r_2 \rvert)\exp[\,\lvert r_1-r_2\rvert-(r_1+r_2)]\} \tag{2}$$
You can download the original paper which I'm trying to reproduce its results here.
The above problem is about equations 17 and 18.
So you are trying to do $$\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}f(r_2,\theta_2,\phi_2)r_2^2\sin\theta_2\,d\phi_2\,d\theta_2\,dr_2$$ I am assuming that $\theta$ is the angle between $\vec r_1$ and $\vec r_2$ so that $\vec r_1\cdot\vec r_2=r_1r_2\cos\theta$. Our first transformation is to rotate the $\vec r_2$ coordinate system so that the $z_2$-axis is parallel to $\vec r_1$. The only change visible in our integral is that now $\cos\theta=\cos\theta_2$ which makes the integral much simpler.
The $\phi_2$ integral is $$\int_0^{2\pi}d\phi_2=2\pi$$ If we let $u=r_1^2+r_2^2-2r_1r_2\cos\theta_2$ then $(r_1-r_2)^2\le u\le(r_1+r_2)^2$, $du=2r_1r_2\sin\theta_2d\theta_2$ and when $\theta_2=0$, $u=(r_1-r_2)^2=u_-$ and when $\theta_2=\pi$, $u=(r_1+r_2)^2=u_+$. The $\theta_2$ integral is $$\begin{align}\int_{u_-}^{u_+}e^{\sqrt u}\frac{du}{2r_1r_2}&=\int_0^{\pi}e^{\sqrt{r_1^2+r_2^2-2r_1r_2\cos\theta_2}}\sin\theta_2\,d\theta_2\\ &=\frac1{r_1r_2}\int_{\sqrt{u_-}}^{\sqrt{u_+}}e^v\cdot v\,dv\\ &=\left.\frac1{r_1r_2}(v-1)e^v\right|_{\sqrt{u_-}}^{\sqrt{u_+}}\\ &=\frac1{r_1r_2}\left[(|r_1+r_2|-1)e^{|r_1+r_2|}-(|r_1-r_2|-1)e^{|r_1-r_2|}\right]\\ &=\frac{e^{r_1+r_2}}{r_1r_2}\left[r_1+r_2-1-(|r_1-r_2|-1)e^{|r_1-r_2|-r_1-r_2}\right]\end{align}$$ Where we have also used the substitution $u=v^2$ so that $du=2v\,dv$ and when $u=u_-=(r_1-r_2)^2$, $v=\sqrt{u_-}=|r_1-r_2|$ and when $u=u_+$, $v=\sqrt{u_+}=|r_1+r_2|=r_1+r_2$. So your result is $$2\pi\frac{e^{r_1}}{r_1}\int_0^{\infty}r_2e^{r_2}\left[r_1+r_2-1-(|r_1-r_2|-1)e^{|r_1-r_2|-r_1-r_2}\right]dr_2$$ Seems to match the problem statement.