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?
 A: Note that  $f(x) =\ln|1-x|-\ln(1+x^2)$. but we know that 
$$ (\ln|1-x| )' =-\frac{1}{1-x} = -\sum_{n=1}^{\infty} x^n ~~~~\mbox{only for}~~0\le |x|<1\\\implies \ln|1-x|  = -\sum_{n=1}^{\infty}\frac{x^n}{n} ~~~~\mbox{only for}~~0\le |x|<1$$
 also from since $-\ln(1-h) = \sum_{n=1}^{\infty}\frac{h^n}{n} ~~~~\mbox{only for}~~0\le h<1$ we have 
$$\ln(1+x^2) = -\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{n} ~~~~\mbox{only for}~~0\le |x|<1$$
hence 
$$f(x) = \sum_{n=1}^{\infty}\frac{ (-1)^{n}x^{2n}-x^n }{n} ~~~~\mbox{only for}~~0\le |x|<1$$
A: So you have $$f(x)=\ln {\frac{|1-x|}{1+x^2}},$$ that means
$$f'(x)=-\frac1{1-x}-\frac{2x}{1+x^2}=(-1-x-x^2-\ldots)-(2x-2x^3\pm\ldots)=-1-3x-x^2+\ldots$$
Integration gives $$f(x)=-x-\frac32x^2-\frac13x^3+\ldots,$$
since $f(0)=0$.
A: 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
$$
a_n=\begin{cases}
-\dfrac{1}{(n-1)/2} & \text{$n$ odd} \\[8px]
-\dfrac{1}{n}-\dfrac{(-1)^{n/2}}{n/2} & \text{$n$ even}
\end{cases}
$$
