2.196. $\frac{\left|x^{2}-1\right|+x^{2}}{2 x^{2}-1}-\frac{|x-1|}{x-1}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq \pm \frac{\sqrt{2}}{2}, \\ x \neq 1 .\end{array}\right.$

Expanding the absolute values with consideration of the domain of definition, we consider three cases:

1) $\left\{\begin{array}{l}x \in(-\infty ;-1), \\ \frac{x^{2}-1+x^{2}}{2 x^{2}-1}+\frac{x-1}{x-1}=\frac{2 x^{2}-1}{2 x^{2}-1}+1=1+1=2 ;\end{array}\right.$
2) $\left\{\begin{array}{l}x \in\left[-1 ;-\frac{\sqrt{2}}{2}\right) \cup\left(-\frac{\sqrt{2}}{2} ; \frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2} ; 1\right), \\ \frac{-\left(x^{2}-1\right)+x^{2}}{2 x^{2}-1}+\frac{x-1}{x-1}=\frac{1}{2 x^{2}-1}+1=\frac{1+2 x^{2}}{2 x^{2}-1}=\frac{2 x^{2}}{2 x^{2}-1}\end{array}\right.$
3) $\left\{\begin{array}{l}x \in(1 ; \infty), \\ \frac{x^{2}-1+x^{2}}{2 x^{2}-1}-\frac{x-1}{x-1}=\frac{2 x^{2}-1}{2 x^{2}-1}-1=1-1=0 \text {. }\end{array}\right.$

Answer: 2, if $x \in(-\infty ;-1) ; \frac{2 x^{2}}{2 x^{2}-1}$, if $x \in\left[-1 ;-\frac{\sqrt{2}}{2}\right) \cup\left(-\frac{\sqrt{2}}{2} ; \frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2} ; 1\right) ; 0$, if $x \in(1 ; \infty)$.