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?

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 $f(0)=0$.
