<s>[INST] Taylor series expansion of the function $f(x)=x \arctan x-0.5 \log(1+x^2)$ about the origin int the region {$|x|\le1$} Find the Taylor series expansion of the function

$\color {green}{f(x)=x \tan^{-1} x-0.5 \log(1+x^2)}$ about the origin int the region {$|x|\le1$}

My effort: I know $\displaystyle \log (1+x)=x-x^2/2+x^3/3-x^4/4...$ now replace $\displaystyle x$ by $\displaystyle x^2$ then we get
$ \displaystyle \frac{1}{2} \log (1+x^2)=x^2/2-x^4/4+x^6/6-x^8/8+....$  (1)
and 
$\displaystyle x \arctan x=x\int (1+x^2)^{-1/2}dx= x^2-x^4/6+3x^5/32 -....$ (2), 
by (2)-(1) I have the answer which does not match with the answer of the book which is$$ \sum_{0}^{\infty}\frac{(-1)^{n-1}x^{2n}}{2n(2n-1)}.$$
My question is my process wrong? Or there is a miscalculation in my process? Is there any better approach to solve this kind of problem? Please give a answer.
 [/INST] it is easier to find the maclaurin series for 
$$f^\prime(x) = \tan^{-1}(x)  = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} + \cdots$$ and integrate this series alsin rthe fact that $f(0) = 0$  which gives you 
$$f(x) = x\tan^{-1}x - 0.5 \ln{(1+x^2)} = \frac{x^2}{2} - \dfrac{x^4}{3.4} + \dfrac{x^6}{5.6} + \cdots$$ 
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