5. Given $f(x)=\frac{\ln \left(x^{2}+1\right)}{(x+1)+3}$, then $y=f(3-x)$ and $y=f(3+x)$ are symmetric about the line ).
A. $x=-3$
B. $x=0$
C. $x=3$
D. $x=6$
5. B

Let $\left(x_{0}, y_{0}\right)$ be on $y=f(3-x)$, then the point $\left(x_{0}, y_{0}\right)$ symmetric to the line $x=a$ is $(2 a$ $\left.-x_{0}, y_{0}\right)$, which must be on $y=f(3+x)$, i.e., $y_{0}=f\left(3+2 a-x_{0}\right)$. Also, $y_{0}=f\left(3-x_{0}\right)$. It is easy to see that $f(x)$ is not a periodic function, so $2 a=0$. Thus, $a=0$.