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. 

$$
\begin{align}
x
&=\int\frac{\,\mathrm{d}y}{y^2-4}\\
&=\frac14\int\left(\frac1{y-2}-\frac1{y+2}\right)\,\mathrm{d}y\\
&=\frac14\log\left(\frac{y-2}{y+2}\right)+C
\end{align}
$$
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.
