1464. Expand $\ln \frac{1+x}{1-x}$ into a power series in $x$.
Solution. $\ln \frac{1+x}{1-x}=\ln (1+x)-\ln (1-x)$.

We already know that

$$
\ln (1+x)=\frac{x}{1}-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots(-1<x \leqslant 1)
$$

Therefore,

$\ln (1-x)=-\frac{x}{1}-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\ldots(-1 \leqslant x<1)$.

Subtracting the second series term by term from the first, we get $\ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots+\frac{x^{2 n-1}}{2 n-1}+\ldots\right)(-1<x<1)$.