$egingroup$ The logarithm is a function, definition that it has a well defined value for a offered $x$. Girlfriend can't leaving an undetermined constant in the an interpretation ! $endgroup$
Because it is not true that we have$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots+C$$for an arbitrary constant $C$. Since, when $x=0$, the LHS is $0$ and RHS is $C$, $C=0$.
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Since the original duty is $log (1+x)$ and for $x=0$ we have actually $log (1+0)=0$ we need that likewise the series is zero for $x=0$.
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