Consider the differential equation with $\frac{dx}{dt}=1-x$ and $x=0$ when $t=0$.
The answer uses the result that $\int \frac{dx}{1-x}=\ln (1-x)$, hence getting the solution $x=1-e^{-t}$.
However I use $\int \frac{dx}{1-x}=\ln \left|1-x\right|$ instead, which gets me $x=1-e^{-t} \text{ or }1+e^{-t}$.
Am I right, or is it standard to not use the absolute signs?