<s>[INST] What's the simplest way to solve $\int \frac{1}{4-v^2}dv$? I did a substitution on a DE and ended up with this:
$\int \frac{1}{4-v^2}dv$
I tried a trig substitute but things got a bit hairy.  WolframAlpha recommended a far less intuitive substitution.  I can't remember what it was.
What is the simplest way to solve this integral?
 [/INST] I think partial fractions may indeed be the "simplest" way, but you mentioned trigonometric substitution getting a bit hairy, and I figured I would give my best derivation. It is a useful skill. This would be advantageous speed wise only if you knew the integral of the secant function from memory.
Consider the more general integral,
$$\int\frac{1}{a^2-v^2} \, dv.$$
If we let 
$$v=a\sin\theta,$$
then
$$dv=a\cos \theta \, d \theta,$$
and
$$a^2-v^2=a^2-a^2\sin^2 \theta=a^2\left( 1-\sin^2 \theta \right)=a^2 \cos^2 \theta.$$
We substitute to get
$$
\begin{align*}
\int \frac{1}{a^2-v^2} \, dv &= \int \frac{a \cos \theta \, d \theta}{a^2 \cos^2 \theta} \\
&=\frac{1}{a}\int\sec \theta \, d \theta \\
&=\frac{1}{a}\ln |\sec \theta + \tan \theta|+c.
\end{align*}
$$
For the back substitution, 
$$v=a \sin \theta \Rightarrow \sin \theta = \frac{v}{a}.$$
We form right a triangle with angle $\theta$, side $v$ opposite $\theta$, and hypotenuse $a$. Thus the side adjacent $\theta$ has length $\sqrt{a^2-v^2}$. Reading directly from the triangle,
$$\frac{1}{a}\ln |\sec \theta + \tan \theta|+c=\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c.$$
We are done, but this certainly cleans up well (like in an integral table).
$$
\begin{align*}
&\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c \\
=& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a^2-v^2}} \right|+c \\
=& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a+v}\sqrt{a-v}} \right|+c \\
=&\frac{1}{a}\ln \left| \frac{\sqrt{a+v}}{\sqrt{a-v}} \right|+c \\
=&\frac{1}{a}\ln \left| \left( \frac{a+v}{a-v} \right)^{\frac{1}{2}} \right|+c \\
=&\frac{1}{2a}\ln \left| \frac{a+v}{a-v}  \right|+c. \\
\end{align*}
$$
Thus your particular integral is
$$\int \frac{1}{4-v^2} \, dv=\frac{1}{4}\ln \left| \frac{a+v}{a-v} \right|+c.$$
It might not seem like the "easiest" way, but it is certainly doable, and in my opinion, fun.
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