2. Given $f(x)=2^{x}+\ln \left(\sqrt{x^{2}+1}+x\right)-2^{-x}+1011$, then the solution set of the inequality $f(2 x+1)+f(x)>2022$ is
$\qquad$ .
Given that $g(x)=f(x)-1011=2^{x}-2^{-x}+\ln \left(\sqrt{x^{2}+1}+x\right)$ is an odd function and is increasing on $\mathbf{R}$, then
$$
\begin{array}{l}
f(2 x+1)+f(x)>2022 \Rightarrow f(2 x+1)-1011>1011-f(x) \\
\Rightarrow g(2 x+1)>-g(x)=g(-x) \Rightarrow 2 x+1>-x \Rightarrow x>-\frac{1}{3},
\end{array}
$$

Therefore, the solution set of the inequality $f(2 x+1)+f(x)>2022$ is $\left(-\frac{1}{3},+\infty\right)$.