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 can set $\sqrt{a+x^2}=x+t$, so $a+x^2=x^2+2tx+t^2$ and
$$
x=\frac{a-t^2}{2t}=\frac{a}{2t}-\frac{t}{2},
\qquad
dx=\left(-\frac{a}{2t^2}-\frac{1}{2}\right)dt=
-\frac{a+t^2}{2t^2}\,dt
$$
Moreover
$$
\sqrt{a+x^2}=\frac{a-t^2}{2t}+t=\frac{a+t^2}{2t}
$$
Hence the integral becomes
$$
-\int\frac{(a+t^2)^2}{4t^3}\,dt=
-\frac{1}{4}\int\left(\frac{a^2}{t^3}+\frac{2a}{t}+t\right)dt
$$
which is elementary.
