<s>[INST] 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!)
 [/INST] Knowing the answer, as i mentioned it in the comments, the substitution in
the answer of J.G. has best chances to work, and it works in a line.
Alternatively...
Starting from the last expression in the OP, we may use the Euler substitution
$$
t = \sqrt{\frac{u+4}u}=\sqrt{1+ \frac 4u}\ .
$$
Then we have formally successively $t^2=1+\frac 4u$, $t^2-1=\frac 4u$,
$u=\frac 4{t^2-1}$,
$du=-\frac{8t}{(t^2-1)^2}\; dt$.
The substitution from $x$ to $t$ is thus $\color{blue}{t=\frac{\sqrt{x^2+4}}x}$. (Assuming $x>0$.)
Then with the used substitutions
$$
\begin{aligned}
\int\frac{dx}{\sqrt{x^2+4}}
&=\frac12\int\frac{du}{\sqrt{u(u+4)}}
=\frac12\int\frac{du}{u\sqrt{\frac{u+4}u}}\\
&=\frac12\int\frac{-\frac{8t}{(t^2-1)^2}\; dt}{\frac 4{t^2-1}\cdot t}
=-\int\frac{dt}{t^2-1}\\
&=
\frac 12\log\frac{t+1}{t-1}+C
=
\frac12
\log
\frac
{\color{blue}{\frac{\sqrt{x^2+4}}x}+1}
{\color{blue}{\frac{\sqrt{x^2+4}}x}-1}+C
\\
&=\log(\sqrt{x^2+4}+x)+C'\ .
\end{aligned}
$$
(At the last point we note that $(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)$ is a constant.)
</s>