Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution? The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:
$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)

$$\frac 1{\sqrt{x^2+4}}=\frac 14\frac{(x^2+4)-x^2}{\sqrt{x^2+4}}=\frac 14\left(\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}\right)$$
$\int\sqrt{x^2+a^2}$ can be computed using IBP without any trig sub.
For the second part, consider the sub $t:=\sqrt{x^2+4}$ so that $t^2=x^2+4$ and $t\,\mathrm dt=x\,\mathrm dx$
$$\int\frac{x^2}{\sqrt{x^2+4}}\,\mathrm dx=\int\frac{t^2-4}{t}\frac t{\sqrt{t^2-4}}\,\mathrm dt=\int\sqrt{t^2-4}\,\mathrm dt$$
which is again computable using IBP without any trig sub.
