Triple Integrals in Spherical Coordinates where (z-2)^2 $$\iiint \frac{1}{\sqrt{x^2+y^2+(z-2)^2}}$$
for $x^2+y^2+z^2 = 1$
I've used spherical coordinates, like this:
$x=\rho sin\phi cos\theta$; $y=\rho sin\phi sin\theta$; $z=\rho cos\phi$ and $J=\rho^2 sin\phi$
$$\rho^2 sin^2\phi cos^2\theta + \rho^2 sin^2\phi sin^2\theta +\rho^2 cos^2\phi = 1$$
$$\rho^2 sin^2\phi + \rho^2 cos^2\phi = 1$$
$$\rho^2 = 1$$
$$\rho = 1$$
and when i put coordinates in my initial integral and simplify it, i get this:
$$\int_0^{2\pi}\int_0^\pi\int_0^1\frac{1}{\sqrt{\rho^2-4\rho cos\phi +4}}\rho^2sin\phi d\theta d\phi d\rho$$
and i can not move from there. 
Please, help.

I think it would be better if you first use $t=z-2$. This way, you have $(x^2+y^2+z^2=1)\rightarrow (x^2+y^2+t^2+4t+3=0)$. Then, use the spherical coordinates. Now, your integral would be really simple, and you need to find the proper boundaries for new integral. 
$(x^2+y^2+t^2+4t+3=0) \rightarrow \rho^2+4\rho cos\phi +3=0 \rightarrow \rho =-2cos\phi \pm \sqrt{4(cos\phi)^2-3} \rightarrow -2cos\phi - \sqrt{4(cos\phi)^2-3}\leq \rho\leq  -2cos\phi + \sqrt{4(cos\phi)^2-3}$.
