Can someone explain the integration of $\sqrt{v²+\tfrac14}$ to me? I am currently trying to integrate this root:
$$\sqrt{v^2+\frac{1}{4}}$$
According to several integration calculators on the web it is:
$$\frac{\operatorname{arsinh}(2v)}{8} +\frac{v\sqrt{v^2+\tfrac{1}{4}}}{2}$$
However, I just can't get my head around it. I have absolutely no idea where that $\operatorname{arsinh}$ is coming from.
My take on this would have been:
$$f = \left(v^2+\frac{1}{4}\right)^{\frac{1}{2}}\implies F = \frac{2}{3}\cdot\frac{1}{2v}\left(v^2+\frac{1}{4}\right)^{\frac{3}{2}}$$

People don't seem comfortable going directly to the hyperbolic trig functions, but that is simplest here. 
$$   v = \frac{1}{2} \sinh t  $$
$$   v^2 + \frac{1}{4} = \frac{1}{4} \cosh^2 t  $$
$$   \sqrt{v^2 + \frac{1}{4}} = \frac{1}{2} \cosh t  $$
$$  dv = \frac{1}{2} \cosh t \; dt $$
$$ \color{blue}{  \sqrt{v^2 + \frac{1}{4}} \; dv = \frac{1}{4} \cosh^2 t \; dt} $$
Next we want the double "angle" formulas. In general,
$$ \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y   $$
$$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y   $$
so
$$ \cosh 2 t = \cosh^2 t + \sinh^2 t = 2 \cosh^2 t - 1, $$ and
$$ \color{red}{ \cosh^2 t = \frac{1 + \cosh 2t}{2}}.  $$
$$ \color{magenta}{  \sqrt{v^2 + \frac{1}{4}} \; dv =  \frac{1 + \cosh 2t}{8} \; dt} $$
The integral $dt$ is 
$$ \color{green}{ \frac{t}{8} + \frac{\sinh 2t}{16}.}  $$
We have $$ t = \operatorname{arsinh} 2v.  $$ We already found
$$  \cosh t = 2 \sqrt{v^2 + \frac{1}{4}}  $$ and
$$  \sinh t = 2 v, $$ whence
$$ \sinh 2t = 2 \sinh t \cosh t = 8 v \, \sqrt{v^2 + \frac{1}{4}}  $$
Put them together, the integral $dv$ is
$$ \color{green}{ \frac{ \operatorname{arsinh} 2v}{8} + \frac{ v \, \sqrt{v^2 + \frac{1}{4}}}{2}.}  $$
I looked on wikipedia, they say there is some preference for arsinh over both arcsinh and argsinh. News to me.
