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.
 A: 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
$$\int_0^1 \frac{r^2}{\sqrt{2 + r^2}} dr = \left[  \frac{1}{2} r \sqrt{r^2 + 2} - \ln(\sqrt{2 + r^2} + r) \right]_0^1 = \ ...$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $\ds{\root{2 + r^{2}} - r=t\ \imp\ r = \frac{1}{t} - \half\,t}$ such that:
\begin{align}&\color{#66f}{\large\int_{0}^{1}\frac{r^{2}}{\root{2 + r^{2}}}\,\dd r}
=\int^{\root{3}- 1}_{\root{2}}\pars{\frac{1}{t} - \frac{1}{4}\,t - \frac{1}{t^{3}}}
\,\dd t
=\color{#66f}{\large\half\,\bracks{\root{3} + \ln\pars{2 - \root{3}}}}
\\[5mm]&\approx{\tt 0.2075}
\end{align}
A: Integration by WA gives
$$
\int \frac{r^2}{\sqrt{2+r^2}} \,dr 
=
\frac{1}{2}r\sqrt{r^2+2} - \sinh^{-1}\left( \frac{r}{\sqrt{2}}\right)+ C
$$
