*specific*continuous $C$ and not simply leave at as an arbitrarily constant? and why carry out we discover the specific continuous we need by setup x=0 and also solve the given equation?

$egingroup$ The logarithm is a function, meaning that it has actually a well characterized value for a given $x$. You can't leave an undetermined consistent in the meaning ! $endgroup$

Because the is

*not*true that we have$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots+C$$for one

*arbitrary*consistent $C$. Since, when $x=0$, the LHS is $0$ and RHS is $C$, $C=0$.

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Since the original function is $log (1+x)$ and also for $x=0$ we have actually $log (1+0)=0$ we need that also the collection is zero because that $x=0$.

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