How to I find the Taylor series of $\ln {\frac{|1-x|}{1+x^2}}$? 
The expression is $$\ln {\frac{|1-x|}{1+x^2}}$$

I'm told there's an easy way to do it to get the first 2 non-zero term but I ended up differentiating this answer several times and got a very long answer that is not correct.
What I did in specific:

*

*split up the ln expressions.


*Differentiate once, so I get 1/|1-x| and 2x/(1+x^2).


*Differentiate another time, and get an even longer expression.
What is the easy way to do this that I do not see?

If you mean the Taylor series at $x=0$, then you can rewrite your function as
$$
\ln(1-x)-\ln(1+x^2)=
-\sum_{n\ge1}\frac{x^n}{n}-\sum_{n\ge1}\frac{(-1)^nx^{2n}}{n}
=\sum_{n\ge1}a_nx^n
$$
where
$$
a_n=\begin{cases}
-\dfrac{1}{(n-1)/2} & \text{$n$ odd} \\[8px]
-\dfrac{1}{n}-\dfrac{(-1)^{n/2}}{n/2} & \text{$n$ even}
\end{cases}
$$
