2.224. $\frac{1}{\sqrt{x^{2}+4 x+4}}+|x-2|$.
## Solution.

Domain of definition: $x \neq-2$.

$$
\begin{aligned}
& \frac{1}{\sqrt{x^{2}+4 x+4}}+|x-2|=\frac{1}{\sqrt{(x+2)^{2}}}+|x-2|=\frac{1}{|x+2|}+|x-2|= \\
& =\frac{1+|x+2| \cdot|x-2|}{|x+2|}=\frac{1+|(x+2)(x-2)|}{|x+2|}=\frac{1+\left|x^{2}-4\right|}{|x+2|}= \\
& =\left\{\begin{array}{l}
\frac{1+x^{2}-4}{-(x+2)}=\frac{x^{2}-3}{-(x+2)}=\frac{3-x^{2}}{x+2} \text { for } x<-2 ; \\
\frac{1-\left(x^{2}-4\right)}{x+2}=\frac{5-x^{2}}{x+2} \text { for }-2<x<2 ; \\
\frac{1+x^{2}-4}{x+2}=\frac{x^{2}-3}{x+2} \text { for } x \geq 2 .
\end{array}\right.
\end{aligned}
$$

Answer: $\frac{3-x^{2}}{x+2}$, if $x \in(-\infty ;-2)$; $\frac{5-x^{2}}{x+2}$, if $x \in(-2 ; 2)$; $\frac{x^{2}-3}{x+2}$, if $x \in[2 ; \infty)$