<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] $$
\begin{aligned}
\int \frac{d x}{\sqrt{x^{2}+4}} &\stackrel{y=\frac{1}{x} }{=}-\int \frac{d y}{y \sqrt{1+4 y^{2}}} \\
&=-\frac{1}{4} \int \frac{d (\sqrt{1+4 y^{2}})}{y^{2}} \\
&=-\int \frac{d (\sqrt{1+4 y^{2}})}{\left(\sqrt{1+4 y^{2}}\right)^{2}-1} \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{1+4 y^{2}}+1}{\sqrt{1+4 y^{2}}-1}\right| \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{x^{2}+4}+x}{\sqrt{x^{2}+4}-x}\right|+c \\
&=\ln \left|\sqrt{x^{2}+4}+x\right|+C
\end{aligned}
$$
</s>