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

What properties of
$\log$
can you use break down$\ln\!\left(\frac{1x}{1+x^2}\right)$
? 
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$\ln\!\left(\frac{1x}{1+x^2}\right) = \ln(1x)\ln(1+x^2)=$
$f(x)f(x^2).$
Since the Taylor expansion of$f(x)$
is given, substitute$x$
and$x^2$
to obtain the terms up to$x^4$
:$$ \begin{aligned} \ln\!\left(\frac{1x}{1+x^2}\right) &=(x\frac{x^2}{2}\frac{x^3}{3}\frac{x^4}{4}\ldots) \\ &\qquad(x^2\frac{x^4}{2}+\ldots)\\ &=x\frac{3x^2}{2}\frac{x^3}{3}+\frac{x^4}{4}+\ldots \end{aligned} $$