<s>[INST] 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:
$$f = \left(v^2+\frac{1}{4}\right)^{\frac{1}{2}}\implies F = \frac{2}{3}\cdot\frac{1}{2v}\left(v^2+\frac{1}{4}\right)^{\frac{3}{2}}$$
 [/INST] 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. 
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