The Taylor expansion of $\ln(1+x)$ is defined as $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$. Expand $\ln\!\left(\frac{1-x}{1+x^2}\right)$ up to and including the 4th power of $x$.

What properties of $\log$ can you use break down $\ln\!\left(\frac{1-x}{1+x^2}\right)$?

How would you relate each term in the above breakdown to the given identity?

Although this question involves Taylor expansion, we do not need to know how to formally expand a function into a Taylor series to answer it. Let $f(x)=\ln(1+x)$. Notice that $\ln\!\left(\frac{1-x}{1+x^2}\right) = \ln(1-x)-\ln(1+x^2)=$ $f(-x)-f(x^2).$ Since the Taylor expansion of $f(x)$ is given, substitute $-x$ and $x^2$ to obtain the terms up to $x^4$: \begin{aligned} \ln\!\left(\frac{1-x}{1+x^2}\right) &=(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots) \\ &\qquad-(x^2-\frac{x^4}{2}+\ldots)\\ &=-x-\frac{3x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}+\ldots \end{aligned}