Integral in spherical coordinates, $\Omega$ is the unit sphere, of $\iiint_\Omega 1/(2+z)^2dx\ dy\ dz$ $$\iiint_\Omega \frac{1}{(2+z)^2}dx\ dy\ dz$$

There is a VERY similar question How to integrate $\iiint\limits_\Omega \frac{1}{(1+z)^2} \, dx \, dy \, dz$ here

But this is different.

I like my spherical coordinates to have the angle in the x/z plane taken from "3 oclock" as normal, rather than from 12. So anyway, I got this:

$$\iiint_\Omega \frac{1}{(2+p\sin(\theta))^2}p^2 \cos(\theta)dp\ d\theta\ d\psi$$
Over
$$\Omega = \lbrace(p,\theta,\psi)|\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\psi\in\left[0,2\pi\right],p\in[0,1]\rbrace$$

I'm not sure what to do, I want to do a substitution but I've been explicitly told to use spherical coords, and it'd be a multivariate substitution (involving p and $\psi$)

Thanks.

Addendum:

I'm hoping you guys will make me feel silly and it'll be obvious, but I've been looking at it for a while, I can't see partial fractions helping,I can't get the p out of the denominator, I'm stumped, sadly.

Thoughts:
It actually seems quite easy in Cartesian. Also why can't I substitute, I'm going to have a go, treating it first as integral involving 2+something, then deal with the sin.
$$I=\int^1_0\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{\cos(\theta)p^2}{(2+p\sin(\theta))^2}\int^{2\pi}_0d\psi\ d\theta\ dp$$

$$=2\pi\int^1_0p\int^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{p\cos(\theta)}{(2+p\sin(\theta))^2}d\theta\ dp$$

Let $u=2+p\sin(\theta)$ then $\frac{du}{d\theta}=p\cos(\theta)$

Thus: $$du=p\cos(\theta)d\theta$$
$$\theta=\frac{\pi}{2}\rightarrow u=2+p$$
$$\theta=-\frac{\pi}{2}\rightarrow u=2-p$$

So:

$$I=2\pi\int^1_0p\int^{2+p}_{2-p}\frac{1}{u^2}du\ dp$$
$$=2\pi\int^1_0p\left[\frac{1}{u}\right]^{2-p}_{2+p}dp$$
$$=2\pi\int^1_0p\left(\frac{1}{2-p}-\frac{1}{2+p}\right)dp$$

Integrate this by parts (differentiate p, integrate the brackets)

Tidy up to get:

$$I=4\pi[\ln(3)-1]$$
Doing this in Cartesian would be hellish; the limits would be a nightmare.

It may be easier to write the integral using $u = \sin \theta$ instead.

The resulting integral is

$$2\pi \int_{-1}^1 \int_0^1 \frac{p^2}{(2+pu)^2} \, dp \, du$$

A table of integrals helps here.  Consulting wikipedia tells us that

$$\int \frac{x^2}{(ax+b)^2} \, dx = \frac{1}{a^3} \left( ax -2b \ln (ax + b) - \frac{b^2}{ax+b} \right) + C$$

That helps us take care of the radial integral:

$$\int_0^1 \frac{p^2}{(2+pu)^2} \, dp = \frac{1}{u^3} \left(u - 0 - 4 \ln (u+2) + 4 \ln 2 - \frac{4}{u+2} + \frac{4}{2} \right)$$

You can probably finish the problem from here.