Consider $\dot{x}=4x^{2}-16$. I am solving the ODE above, it is a question from Strogatz Nonlinear dynamics and chaos, chapter 2 question 2.2.1.
Question
\begin{equation}
\dot{x}=4x^{2}-16
\end{equation}
Answer
\begin{equation}
\frac{\dot{x}}{4x^{2}-16} = 1\\
\frac{\dot{x}}{x^{2}-4} = 4\\
 {{dx\over dt}\over x^2-4}=4 \\
 {{dx\over dt}\over x^2-4}. dt=4. dt  \\
 {dx\over x^2-4}=4dt \\
\int \frac{1}{x^{2}-4} dx = \int 4 dt \\
\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} \\
x = 2 \frac{1 + C_{2}e^{16t}}{1 - C_{2}e^{16t}}
\end{equation}
\begin{equation}
C_{2}(t=0) = \frac{x-2}{x+2}
\end{equation}
Summary
I am looking to understand the intermediary step in the proof above. How do we get to this step $\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} $ from the previous step. Can we remove the constant $\frac{1}{4} $ then integrate the remaining portion?

The author has used the method of "partial fractions" to write
$$
\frac{1}{x^2-4} = \frac{A}{x-2} + \frac{B}{x+2},
$$
(solving for $A$ and $B$, which I'm not going to do), integrated each of the right hand items to get a "log" term, and then combined a difference-of-logs into a log-of-a-quotient. There's nothing subtle here except skipping steps from calculus class.
