I am not sure whether I am doing it correctly. So,
$$\frac{2+x}{x^2+2x+2} = \frac{2}{x^2+2x+2} + \frac{x}{x^2+2x+2} = F_1 + F_2,$$ $$x^2+2x+2 = (x - x_1)(x-x_2), \text{where} \\ x_1 = i+1, \\x_2 = i -1;$$ $$F_1 = \frac{2}{(x - x_1)(x-x_2)} = \frac{A}{(x - x_1)} + \frac{B}{(x - x_2)} = \frac{1}{(x - i - 1)} - \frac{1}{(x - i + 1)}$$
Taking the first one: $$ \frac{1}{(x - i - 1)} = \frac{1}{x + 1 - i - 2} = \frac{-1}{i+2}\cdot\frac{1}{1 - \dfrac{x+1}{i+2}} = \frac{-1}{i+2} \sum\limits_{n = 0}^{\infty}\frac{(x+1)^n}{(i+2)^n} = \\ = -\sum\limits_{n = 0}^{\infty}\frac{(x+1)^n}{(i+2)^{n+1}}$$ Then the second therm, combining, then $F_2$, combining all together. Is that correct way? Just my answer looks ugly, not like wolfram alpha shows.
Let $u=x+1$, then $$\frac{2+x}{x^2+2x+2}=\frac{1+u}{1+u^2}=(1+u)(1-u^2+u^4-u^6+\cdots)$$ $$=1+u-u^2-u^3+u^4+u^5-u^6-u^7+\cdots$$ $$=1+(x+1)-(x+1)^2-(x+1)^3+(x+1)^4+\cdots.$$