How many arbitrary constants to put in these integrals?

This is a good question.

1. If \begin{align}g_1(x)&:= \begin{cases} \log|x|+C_1, &x<0;\\ \log|x|+C_2, &x>0\end{cases}\\ g_2(x)&:= \log|x|+C\quad(x\ne0),\end{align} then $$g_1'(x)\equiv\dfrac1x\equiv g_2'(x),$$ but strictly speaking, $$g_1(x)=\int \frac{1}{x} \,\mathrm dx\ne g_2(x).$$ $\int \frac{1}{x} \,\mathrm dx\ne g_2(x)$ because the integrand's domain is not a single interval and thus the indefinite integral has independent integration constants.
2. On the other hand, if \begin{align}h_1(x)&:= \begin{cases} \frac12x|x|+C_1, &x<0;\\ \frac12x|x|+C_2, &x\geq0 \end{cases}\\ h_2(x)&:= \frac12x|x|+C,\end{align} then $$h_1'(x)\not\equiv|x|\equiv h_2'(x),$$ and $$h_1(x)\ne\int |x| \,\mathrm dx = h_2(x).$$ $h_1'(x)\not\equiv|x|$ because $h_1$ is not generally even continuous at $0.$

Here is another indefinite integral; this time, the integration constant in case 1 depends on that in case 2 but is independent of that in case 3: $$\int\frac{|x+1|}{x}\,\mathrm dx =\begin{cases} -x-\ln(-x)\color{red}{-2+C_1}, &x<-1;\\ x+\ln(-x)\color{red}{+C_1}, &-1\le x<0;\\ x+\ln(x)\color{red}{+C_2}, &x>0. \end{cases}$$