Analytical evaluation of double integral I'm looking to evaluate the following integral:
$$\int^a_{-a}\int^a_{-a}\frac{\sqrt{x^2+y^2}}{4a^2} \,\mathrm{d}x \mathrm{d}y,$$
where $a> 0$.
Mathematica gives the following answer:
$$ \frac{a}{3} (\sqrt{2} + \sinh^{-1}(1)).$$
This goes well beyond my basic calculus 101 training.
Is anybody able to give me a step-by-step analytical solution?
Regards

HINTS:
Exploiting the even symmetry and making use of 
$$\int \sqrt{x^2+y^2}\,dx=\frac x2\sqrt{x^2+y^2}+\frac{x^2}{2} \log(x+\sqrt{x^2+y^2})+C$$
evaluation of the inner integral yields
$$\begin{align}
\int_{-a}^a\int_{-a}^a \frac{\sqrt{x^2+y^2}}{4a^2}\,dx\,dy &= \frac1{a^2} \int_0^a \int_0^a \sqrt{x^2+y^2}\,dx\,dy \\\\
&=\frac1{a^2}\int_0^a\left(\frac a2\sqrt{a^2+y^2}+\frac{a^2}{2} \log(a+\sqrt{a^2+y^2})-\frac{a^2}{2}\log(y)\right) \,dy
\tag 1
\end{align}$$
Use integration by parts for the middle term in the integral on the right-hand side of $(1)$


EDIT:
The OP has requested to see a presentation on evaluation of $\displaystyle \int_0^a \log(a+\sqrt{y^2+a^2})\,dy$ by integration of parts. To that end, we proceed.

Let $u=\log(a+\sqrt{y^2+a^2})$ and $v=y$.  Then, we can write
$$\begin{align}
\int_0^a \log(a+\sqrt{y^2+a^2})\,dy&=a\log(a(1+\sqrt{2}))-\int_0^a \frac{y^2}{a+\sqrt{y^2+a^2}}\frac{1}{\sqrt{y^2+a^2}}\,dy\\\\
&=a\log(a(1+\sqrt{2}))-\int_0^a \frac{\sqrt{y^2+a^2}-a}{\sqrt{y^2+a^2}}\,dy\\\\
&=a\log(a(1+\sqrt{2}))-a+a\int_0^a \frac{1}{\sqrt{y^2+a^2}}\,dy\\\\
&=a\log(a(1+\sqrt{2}))-a+a\log(a(1+\sqrt{2}))-a\log(a)\\\\
&=a(\log(a)-1+2\log(1+\sqrt{2}))
\end{align}$$
