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)?

Notice that:
$$\int f(x)dx = \int \dfrac{x}{1-x^2}dx=-\frac{1}{2}\int \frac{du}{u}=-\frac{1}{2}\ln|1-x^2|+C$$
(via the $u$-substitution $u=1-x^2$)
From your note:
$$\ln(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+\dots$$
So substitute $1-x^2$ into $x$ in the natural log taylor expansion, and take the derivative to obtain a series representation of $f(x).$
