Exponential growth/decay formula: what happened to the other constant of integration?

When you integrate both sides, each has a constant - you'd get, for constants $A,B$:

$$ \int{dP\over P}=\int kdt \implies \ln|P|+A = kt+B$$

Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then

$$\ln|P|+A = kt+B \implies \ln|P|=kt+B-A=kt+C$$

This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.

Well, notice that:

$$\ln\left|\text{P}\left(t\right)\right|+\text{C}_1=\text{k}\cdot t+\text{C}_2\tag1$$

Getting $\text{C}_1$ on the other side gives:

$$\ln\left|\text{P}\left(t\right)\right|=\text{k}\cdot t+\text{C}_2-\text{C}_1\tag2$$

But $\text{C}_2-\text{C}_1$ is another constant, so: $$\ln\left|\text{P}\left(t\right)\right|=\text{k}\cdot t+\text{C}\tag3$$