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?

Hint: substitute $$x=\sqrt{a}\sinh(t)$$
