In physics (molecular theory of liquids), the following integral sometimes appears (and don't have proper reference at hand, only a spanish textbook) :
$I(s) = \int_{V_R} f_1(r_1) f_2(r_2) d{\bf x}$
where $V_R$ is a ball of radius $R$ (usually $R=\infty$), $r_1=|\bf{x}|$, $r_2=|{\bf x}-{\bf s}|$ and $s=|{\bf s}|$; WLOG, let ${\bf s}=(0,0,s)$ Changing first to polar coordinates ($x =r_1 \sin \theta \cos \phi$, $y =r_1 \sin \theta \sin \phi$ , $z=r_1 \cos \theta$) and then to bipolar coordinates ($r_1=r_1$, $r_2^2=r_1^2+s^2-2 r_1 s \cos \theta$, $\phi = \phi$) we get:
$I(s) = 2 \pi \int_{0}^{R} \int_{|r_1-s|}^{r1+s}f_1(r_1) f_2(r_2) \frac{r_1 r_2}{s} dr_2 dr_1 $
This is slightly more general than we need, but it seems a interesting/useful formula.
In our case, $f_1 =1$, $f_2(r_2) = 1/r_2$, (and $s=|x|$), hence
$I(s)=2\pi \int_{0}^{R} \int_{|r_1-s|}^{r1+s} \frac{r_1}{s} dr_2 dr_1$
So $\frac{I(s)}{2\pi}= \int_{0}^{s} \frac{r_1}{s} 2 r_1 dr_1 + \int_{s}^{R} \frac{r_1}{s} 2 s dr_1 = \frac{2}{3}s^2 + (R^2 -s^2)$
and
$I(s)=2 \pi \left(R^2-\frac{1}{3}s^2\right)$
This should be valid (unless I've messed something) for ${\bf x}$ inside the ball (i.e. $s \le R$).
Edit: quick and dirty numerical test with Octave/Matlab:
R=5; s=3.5; N=30000; X=rand(N,3)*2*R-R; Y=sqrt(sum(abs(X).^2,2)); X1 = X(Y
It feels right.