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:
$$f = \left(v^2+\frac{1}{4}\right)^{\frac{1}{2}}\implies F = \frac{2}{3}\cdot\frac{1}{2v}\left(v^2+\frac{1}{4}\right)^{\frac{3}{2}}$$
 A: Since $\sqrt{v^2+\frac{1}{4}}$ is of the form $\sqrt{x^2+a^2}$, we can do a trigonometric substitution using $\tan\theta$.
Let $v=\frac{1}{2}\tan u$, $dv=\frac{1}{2}\sec^2(u) du$ 
$$\int \sqrt {v^2+\frac{1}{4}}dv=\int \sqrt {\frac{1}{4}\tan^2 u+\frac{1}{4}}\cdot\frac{1}{2}\sec ^2 (u)du$$
Using the trig identity $1+\tan^2\theta=\sec ^2 \theta$ we have
$$=\frac{1}{4}\int \sec ^3 (u)du$$
Now we can use integration by parts. 
$$ds=\sec ^2(u)du$$
$$s=\tan u$$
$$t=\sec u$$
$$dt=\sec (u) \tan (u) du$$
Then 
$$\int \sec ^3 (u)du=\sec u \tan u-\int \sec (u) \tan^ 2(u)du$$
Using the same identity we can substitute again.
$$=\sec u \tan u - \int \sec u (\sec ^2 u -1)du$$
$$\int \sec^3(u)du=\sec u \tan u -\int \sec ^3 (u)du + \int sec (u) du$$
But $\int \sec ^3 (u)du$ is what we are trying to find. 
$$2\int\sec ^3 (u)du = \sec u \tan u+\int \sec (u) du$$
$$=\sec u \tan u +\ln|\sec u + \tan u |$$
$$\frac{1}{4}\int \sec ^3 (u)du=\frac{1}{8}\sec u \tan u +\frac{1}{8}\ln |\sec u + \tan u|$$
Now we are almost ready to switch back to $v$. We defined $\tan u = 2v$ earlier. If we draw a representative right triangle we can figure out $\sec u$ as well. 

From the triangle it is clear that $\sec u =\sqrt {4v^2+1}$.
Therefore
$$\int\sqrt{v^2+\frac{1}{4}}dv=\frac{1}{4}v\sqrt{4v^2+1}+\frac{1}{8}\ln |2v+\sqrt{4v^2+1}| + C$$
which you can switch to the hyperbolic if you like. 
A: 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.
A: $$\int\sqrt{v^2+\frac{1}{4}}\space\space\text{d}v=$$

Substitute $v=\frac{\tan(u)}{2}$ and $\text{d}v=\frac{\sec^2(u)}{2}\space\space\text{d}u$. Then $\sqrt{v^2+\frac{1}{4}}=\sqrt{\frac{\tan^2(u)}{4}+\frac{1}{4}}=\frac{\sec(u)}{2}$ and $u=\tan^{-1}(2v)$: 

$$\frac{1}{2}\int\frac{\sec^3(u)}{2}\space\space\text{d}u=$$
$$\frac{1}{4}\int\sec^3(u)\space\space\text{d}u=$$
$$\frac{1}{8}\tan(u)\sec(u)+\frac{1}{8}\int\sec(u)\space\space\text{d}u=$$
$$\frac{1}{8}\tan(u)\sec(u)+\frac{1}{8}\int\frac{\sec^2(u)+\sec(u)\tan(u)}{\sec(u)+\tan(u)}\space\space\text{d}u=$$

Substitute $s=\tan(u)+\sec(u)$ and $\text{d}s=(\sec^2(u)+\tan(u)\sec(u))\space\space\text{d}u$: 

$$\frac{1}{8}\tan(u)\sec(u)+\frac{1}{8}\int\frac{1}{s}\space\space\text{d}s=$$
$$\frac{1}{8}\tan(u)\sec(u)+\frac{\ln\left(s\right)}{8}+\text{C}=$$
$$\frac{1}{8}\tan(u)\sec(u)+\frac{\ln\left(\tan(u)+\sec(u)\right)}{8}+\text{C}=$$
$$\frac{1}{8}\tan\left(\tan^{-1}(2v)\right)\sec\left(\tan^{-1}(2v)\right)+\frac{\ln\left(\tan\left(\tan^{-1}(2v)\right)+\sec\left(\tan^{-1}(2v)\right)\right)}{8}+\text{C}=$$
$$\frac{v}{4}\sqrt{4v^2+1}+\frac{1}{8}\ln\left(\sqrt{4v^2+1}+2v\right)+\text{C}=$$
$$\frac{1}{8}\left(2v\sqrt{4v^2+1}+\sinh^{-1}(2v)\right)+\text{C}$$
