<s>[INST] 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?
 [/INST] $$I=\int \frac{dx}{x^{2}-4}$$
Partial fraction decomposition gives us:
$$I=\int \left ( \frac A {x-2}-\dfrac B {x+2} \right)dx$$
$$I=\int \frac{x(A-B)+2(A+B)}{x^{2}-4}dx$$
$$\implies A=B=\dfrac 14$$
$$I=\dfrac 14\int \left ( \frac 1 {x-2}-\dfrac 1 {x+2} \right)dx$$
Then integrate with $\ln $ function .
$$I=\dfrac 14 \ln |{x-2}|-\dfrac  14\ln |{x+2}|+C$$
$$I=\dfrac 14 \ln \left | \dfrac {x-2}{x+2} \right |+C$$
</s>