Finding Taylor Series from existing Series If the Taylor Series of $\ln(x)$ is known:
$$\ln(x) = (x-1) -\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-...$$
Can one find the Taylor series of
$$f(x)= \frac{x}{1-x^2}$$
by manipulating the Taylor series of ln(x)?

As an alternate without using $\ln x$. Make use of the geometric series:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n \implies \frac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n} \implies \frac{x}{1-x^2} = \sum_{n=0}^\infty x^{2n+1} $$
