<s>[INST] Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
 [/INST] $$
x^2 + 2x + 2 = (x+1)^2 + 1
$$
and, with  $\ s = x+1 \ $ you have  
$$
\frac 1{x^2 + 2x + 2} = \frac 1{(x+1)^2 + 1} = \frac 1{s^2 + 1} = \text{D} ( \arctan s ).
$$
Now, just derive the Taylor expansion of  $\ \arctan s \ $ and subsitute $\ s = x +1 $
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