(5) Given the function $f(x)=2016^{x}+\log _{2016}\left(\sqrt{x^{2}+1}+x\right)-2016^{-x}+2$, then the solution set of the inequality $f(3 x+1)+f(x)>4$ with respect to $x$ is ( )
A. $\left(-\frac{1}{2016},+\infty\right)$
B. $\left(-\frac{1}{3},+\infty\right)$
C. $\left(-\frac{1}{2},+\infty\right)$
D. $\left(-\frac{1}{4},+\infty\right)$
Solution 1: Let $g(x)=f(x)-2$, then the function $g(x)$ is an odd function and is increasing over the real numbers,
the inequality is transformed into $g(3 x+1)+g(x)>0 \Leftrightarrow g(3 x+1)>g(-x) \Leftrightarrow 3 x+1>-x \Rightarrow x>-\frac{1}{4}$
Solution 2: $y=f(x)$ is symmetric about the center $(0,2)$ and is increasing over the real numbers $3 x+1+x>0 \Rightarrow x>-\frac{1}{4}$.