$\begingroup$ The logarithm is a function, an interpretation that it has a well defined value because that a given $x$. Friend can't leaving an undetermined continuous in the meaning ! $\endgroup$
Because the is not true that us have$$\log(1+x)=x-\fracx^22+\fracx^33-\fracx^44+\cdots+C$$for an arbitrary continuous $C$. Since, once $x=0$, the LHS is $0$ and also RHS is $C$, $C=0$.
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Since the original function is $\log (1+x)$ and also for $x=0$ we have $\log (1+0)=0$ we need that also the collection is zero because that $x=0$.
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