Solution to $y'=y^2-4$ I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\frac{1}{(y+2)(y-2)} = \frac{A}{y+2}+\frac{B}{y-2}$
Therefore,
$1=Ay-2A+By+2B$.
Since the coefficients must be the same on both sides of the equation it follows that:
$0=A+B$    and $1=-2A+2B$.
Hence, $A=-B$, $B=\frac14$, $A=-\frac14$.
Thus the differential equation should be transformed into:
$-\frac{1}{4} \frac{dy}{y+2} + \frac14 \frac{dy}{y-2} = x+C$
Solving this should yield:
$-\frac14 \ln|y+2| + \frac14 \ln|y-2| = x+C$
which simplifies as:
$\ln(y-2)-\ln(y+2)=4(x+c)$
$\ln[(y-2)/(y+2)]=4(x+c)$
$(y-2)/(y+2)=\exp(4(x+c))$
$y-2=y*\exp(4(x+c)+2\exp(4(x+c))$
$y-y\exp(4(x+c))=2+2\exp(4(x+c))$
$y(1-\exp(4(x+c)))=2(1+\exp(4(x+c)))$
$y= 2(1+\exp(4(x+c)))/(1-\exp(4(x+c)))$
However, when done in Mathematica/Wolfram Alpha the result is given as (proof in the attached image)
$\frac14 \ln(2-y) -\frac14 \ln(2+y) = x + C$
and returns an answer of:
$y= 2(1-\exp(4(x+c)))/(1+\exp(4(x+c)))$.
Can anyone figure out where I have made an error? The only thing I can think of is something with evaluating the absolute values of the natural logarithms. 
 A: $$
\begin{align}
x
&=\int\frac{\,\mathrm{d}y}{y^2-4}\\
&=\frac14\int\left(\frac1{y-2}-\frac1{y+2}\right)\,\mathrm{d}y\\
&=\frac14\log\left(\frac{y-2}{y+2}\right)+C
\end{align}
$$
Therefore,
$$
\frac{y-2}{y+2}=ke^{4x}
$$
or, solving for $y$,
$$
y=2\,\frac{1+ke^{4x}}{1-ke^{4x}}
$$
We get your form of the answer by letting $k\lt0$. It is hard to see this since you need to use a complex $c$ to get the equivalent answer.  Check your answer in the original equation. You'll see that it works.
A: Saturday morning: I do like robjohn's point about $\log |x|$ giving the student some incorrect expectations. What I wrote here is correct and careful, but the same conclusions come from dropping the absolute  value signs and doing the calculations three times, $y < -2,$ $-2 < y < 2,$ $y > 2.$ At some point the vertical asymptotes will be revealed, how solutions with $y > 2$ reach a vertical asymptote and then jump below to $y < -2.$ That is how I usually do things, split into cases early. 
Lost my train of thought. However, if I want $\int \frac{1}{x} dx,$ without absolute values, the answer would be $\log x$ for $x > 0,$ but $\log (-x)$ for $x < 0.$
$-\frac14 \log|y+2| + \frac14 \log|y-2| = x+C$
which simplifies as:
$\log|y-2|-\log|y+2|=4(x+c),$ or
$$ \left| \frac{y-2}{y+2} \right| = k e^{4x}   $$
with $k > 0$ for the nonconstant solutions.
The item inside absolute value signs is a linear fractional or Mobius transformation, once we decide about the $\pm$ signs the inverse is given by the matrix
$$
\left(
\begin{array}{rr}
2 & 2 \\
-1 & 1
\end{array}
\right)
$$
If $y > 2$ or $y < -2,$ those inequalities hold true forever, we have
$$  \frac{y-2}{y+2}  = k e^{4x}   $$
 and we get robjohn's 
$$  y = \frac{2k e^{4x} + 2}{-k e^{4x} + 1}. $$
There is a jump discontinuity: for some value of $x,$ we find that
$-k e^{4x} + 1= 0,$ indeed $e^{4x} = 1/k,$ $4x = - \log k,$ x = $-(1/4) \log k.$ For $x < -(1/4) \log k,$ we have $y > 2.$ There is a vertical asymptote, then for $x > -(1/4) \log k,$ we have $y <  -2.$
In all cases, the curves are asymptotic to the lines $y=2$ and $y=-2.$ Also, note that the ODE is autonomous. Given one of the solutions described, we get another solution by shifting $x$ by any constant we like. If preferred, we can drop the multiplier $k$ by emphasizing the horizontal shifts. 
If, however, we begin with $-2 < y < 2,$ then those inequalites hold forever, we have
$$  \frac{y-2}{y+2}  = - k e^{4x}   $$ and we get the continuous
$$  y = \frac{-2k e^{4x} + 2}{k e^{4x} + 1}. $$
In the picture below, I took $k=1.$ To find all other solutions, replace $x$ by some $x - x_0.$  Put another way, we can take $e^{-4 x_0}= k;$ thus, we can account for all solutions either by varying $k,$ or by deleting $k$ and adjusting $x_0.$    Note also that the part with $y>2$ and the part with $y < -2$ are tied together, the same vertical asymptote. 

