I am trying to evaluate an indefinite integral and this implicit solution came up on integral-calculator.com:
I was wondering how the common denominator of the two terms inside the absolute value argument of the natural log vanished. Is there any explanation to this?
The common denominator didn't exactly "vanish". Instead, note that
$$\begin{equation}\begin{aligned} 2\ln\left(\left|\sqrt{\frac{x^2}{4} + 1} + \frac{x}{2}\right|\right) & = 2\ln\left(\left|\sqrt{\frac{x^2 + 4}{4}} + \frac{x}{2}\right|\right) \\ & = 2\ln\left(\left|\sqrt{\frac{x^2 + 4}{4}} + \frac{x}{2}\right|\right) \\ & = 2\ln\left(\left|\frac{\sqrt{x^2 + 4}}{2} + \frac{x}{2}\right|\right) \\ & = 2\ln\left(\frac{1}{2}\left|\sqrt{x^2 + 4} + x\right|\right) \\ & = 2\left(\ln\left(\frac{1}{2}\right) + \ln\left(\left|\sqrt{x^2 + 4} + x\right|\right)\right) \\ & = 2\ln\left(\left|\sqrt{x^2 + 4} + x\right|\right) - 2\ln(2) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The "C" in the first line is an arbitrary constant, and is not the same value as the $C$ in the second line. If you call the first $C$ a different name, say $C_1$, and the second one $C_2$, you can see from \eqref{eq1A} that
$$C_2 = C_1 - 2\ln(2) \tag{2}\label{eq2A}$$
In other words, the common denominator value, i.e., $-2\ln(2)$, was "absorbed" into the arbitrary constant $C$ in the second line.