How to solve $\int \sqrt{a+x^2} dx $, $a>0$? By setting $a+x^2 = t^2$ I can get $2xdx=2dt$ so $dx=dt/\sqrt{t^2-a}$. But such substitutions only lead to iterate integrals of the form
$$
\int \frac{t^2}{\sqrt{t^2-a}}dt,
$$ 
substituting again $t^2-a=u^2$ we get 
$$
\int udu +\int\frac{2}{\sqrt{u^2+a}}du.
$$
How can I solve the second integral?

You may, alternatively use trigonometric substitution:
In this case, note that $a+ x^2 = (\sqrt a)^2 + x^2$.  
The corresponding trig substitution would be to let $x = \sqrt a\tan \theta\;$ so $\;dx = \sqrt a \sec^{2} \theta\,d\theta = \dfrac {\sqrt a}{\cos^2 \theta}\,d\theta$, and you can continue from there.
A good source to you refer when first practicing trig substitution is this Wikipedia page dedicated to the use of trig substitution for solving many integrals.
