<s>[INST] Triple Integral of $1/\sqrt{2 + x^2 + y^2 + z^2}$ over unit sphere I'm studying triple integrals (physics major), and I'm having trouble solving this little beast:
$$
\iiint_V \frac{1}{\sqrt{2 + x^2 +y^2 +z^2}} \,dx\,dy\,dz$$ 
where V is $$x^2+y^2+z^2=1$$
Of course we use spherical coordinates:
$$I = \iiint_V \frac{r^2 \sin φ}{\sqrt{2+r^2}} \,dr\,dφ\,dθ$$
In order to solve the first integral over r I simplified the denominator using $$\sqrt{2+r^2} = \sqrt{2\left(1+\frac{r^2}{\sqrt{2}^2}\right)}$$
in order to substitute $\tan ω = \frac{r}{\sqrt{2}}$. However again even this integral leads to 2 pages of computations and I still haven't reached a correct result.
Is there any shortcut I'm not seeing?
PS: The limits of $r,θ,φ$ are the standard ones since we are on the unit sphere.
 [/INST] Write
$$\frac{r^2}{\sqrt{a^2 + r^2}} = \sqrt{a^2 + r^2} - \frac{a^2}{\sqrt{a^2 + r^2}}$$
These two terms are now standard integrals, equal respectively to $\displaystyle \frac{1}{2} \left( r \sqrt{a^2 + r^2} + a^2 \ln(\sqrt{a^2 + r^2} + r) \right)$ and $-a^2 \cdot \ln(\sqrt{a^2 + r^2} + r)$
Hence
$$\int_0^1 \frac{r^2}{\sqrt{2 + r^2}} dr = \left[  \frac{1}{2} r \sqrt{r^2 + 2} - \ln(\sqrt{2 + r^2} + r) \right]_0^1 = \ ...$$
</s>