Consider $\dot{x}=4x^{2}-16$.

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.

$$I=\int \frac{dx}{x^{2}-4}$$ 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 . $$I=\dfrac 14 \ln |{x-2}|-\dfrac 14\ln |{x+2}|+C$$ $$I=\dfrac 14 \ln \left | \dfrac {x-2}{x+2} \right |+C$$