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!)
 A: Use an Euler substitution $t=\sqrt{x^2+4}+x$ so $\frac{dt}{dx}=\frac{x}{\sqrt{x^2+4}}+1=\frac{t}{\sqrt{x^2+4}}$ and $\frac{dx}{\sqrt{x^2+4}}=\frac{dt}{t}$.
A: $$\frac 1{\sqrt{x^2+4}}=\frac 14\frac{(x^2+4)-x^2}{\sqrt{x^2+4}}=\frac 14\left(\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}\right)$$
$\int\sqrt{x^2+a^2}$ 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.
A: Substitute $x=\frac{t}{2}-\frac2t $ instead to integrate
$$\int \frac1{\sqrt{4+x^2}}dx = \int \frac1tdt=\ln t+C
$$
A: $$x=2\sinh{z}\Rightarrow dx=2\cosh{z}dz\\
\int\frac{dx}{\sqrt{4+x^{2}}}=\int\frac{2\cosh{z}dz}{2\cosh{z}}=z+c =\sinh^{-1}{(\frac{x}{2})}+c\\
$$
A: Put $$x=2t$$
it becomes
$$\int  \frac{dt}{\sqrt{t^2+1}}=$$
$$\sinh^{-1}(t)+C$$
You can use the fact that if
$$F(t)=\ln(t+\sqrt{t^2+1})$$
then
$$F'(t)=\frac{1}{\sqrt{t^2+1}}$$
A: 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.)
A: $$
\begin{aligned}
\int \frac{d x}{\sqrt{x^{2}+4}} &\stackrel{y=\frac{1}{x} }{=}-\int \frac{d y}{y \sqrt{1+4 y^{2}}} \\
&=-\frac{1}{4} \int \frac{d (\sqrt{1+4 y^{2}})}{y^{2}} \\
&=-\int \frac{d (\sqrt{1+4 y^{2}})}{\left(\sqrt{1+4 y^{2}}\right)^{2}-1} \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{1+4 y^{2}}+1}{\sqrt{1+4 y^{2}}-1}\right| \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{x^{2}+4}+x}{\sqrt{x^{2}+4}-x}\right|+c \\
&=\ln \left|\sqrt{x^{2}+4}+x\right|+C
\end{aligned}
$$
