= Solution
For $|t|<1$, the <geometric series> gives $(1+t)^{-1}=\sum_{n\geq0}(-t)^n$. Its uniform convergence on every closed interval strictly inside $(-1,1)$ permits <termwise integration> from zero to $x$. Therefore
$$
\log(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n},\qquad
\log(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}.
$$
Subtracting cancels every even power and doubles every odd power, giving the <Taylor series>
$$
\boxed{\log\frac{1+x}{1-x}=2\sum_{j=0}^\infty\frac{x^{2j+1}}{2j+1},\qquad |x|<1.}
$$
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