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?
 A: I'm surprised that with all these answers no one has mentioned hyperbolic trigonometric functions. If you know that $\int \frac{1}{1+x^2}dx=\tan^{-1}(x)$ it may not surprise you to learn that $\int \frac{1}{1-x^2} dx=\tanh^{-1}(x)$, and so:
$$\int \frac{1}{4-v^2}dv=\frac{1}{4}\int \frac{1}{1-(\frac{v}{2})^2} dv$$
substitude $u=\frac{v}{2}$ so $2 du=dv$
$$=\frac{1}{2} \int \frac{1}{1-u^2} du=\frac{1}{2} \tanh^{-1}(u)=\frac{1}{2} \tanh^{-1}(\frac{v}{2})$$
To prove the identities, try to mimic the proof that $\int \frac{1}{1+x^2}dx=\tan^{-1}(x)$. That is, put $x=\tan(y)$ to find that $\frac{dy}{dx}=\cos^2(y)=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}$ so that integration with respect to x gives $y=\int \frac{1}{1+x^2}$. You'll need to prove the hyperbolic equivalent of $\sec^2(x)=1+\tan^2(x)$, $\text{sech}^2(x)=1-\tanh^2(x)$, which is trivial using the fundamental property $\cosh^2(x)-\sinh^2(x)=1$ (divide through by $\cosh^2(x)$).
Of course, partial fractions is the way to go if your teacher does not allow this form of an answer. In fact, the partial fraction decomposition and the identity $\int \frac{1}{1-x^2} dx=\tanh^{-1}(x)$ would probably be a sorta neat way to prove that $\tanh^{-1}(x)=\frac{1}{2}(\log(1+x)-log(1-x))$. (Though another method of proving it: writing $y=\tanh(x)=\sinh(x)/\cosh(x)$, expanding into exponentials, and solving for $x$ in terms of $y$ isn't too difficult.)
A: Use partial fractions. Put
$${1\over 4 -x^2} = {A\over 2-x} + {B\over 2+ x}.$$
A: 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.
A: substitute $2z = v,$ and then $z = \sin \theta.$
A: Hint: $\frac{1}{4-u^2}=\frac{\frac{1}{4}}{{2-u}}+\frac{\frac{1}{4}}{{2+u}}$
