2.213. $\frac{\sqrt{1+\left(\frac{x^{2}-1}{2 x}\right)^{2}}}{\left(x^{2}+1\right) \cdot \frac{1}{x}}$.
Solution.

Domain of definition: $\boldsymbol{x} \neq 0$.

$$
\begin{aligned}
& \frac{\sqrt{1+\left(\frac{x^{2}-1}{2 x}\right)^{2}}}{\left(x^{2}+1\right) \cdot \frac{1}{x}}=\frac{\sqrt{1+\frac{x^{4}-2 x^{2}+1}{4 x^{2}}}}{\frac{x^{2}+1}{x}}=\frac{\sqrt{\frac{4 x^{2}+x^{4}-2 x^{2}+1}{4 x^{2}}}}{\frac{x^{2}+1}{x}}=
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0057.jpg?height=234&width=763&top_left_y=1377&top_left_x=226)

$$
\begin{aligned}
& =\left\{\begin{array}{l}
-\frac{x}{2 x}=-\frac{1}{2}, \text { if } x \in(-\infty ; 0) \\
\frac{x}{2 x}=\frac{1}{2}, \text { if } x \in(0 ; \infty)
\end{array}\right.
\end{aligned}
$$

Answer: $-\frac{1}{2}$, if $x \in(-\infty ; 0) ; \frac{1}{2}$, if $x \in(0 ; \infty)$.