5. Let for all real numbers $x$, the function $f(x)$ satisfies $x f(x)=2 f(1-x)+1$.
Then $f(5)=$ $\qquad$
5. $\frac{1}{12}$.

Replacing $x$ with $1-x$ in the given equation yields $(1-x) f(1-x)=2 f(x)+1$.
Eliminating $f(1-x)$ from the two equations gives
$$
f(x)=\frac{x-3}{x^{2}-x+4} \text {. }
$$

Thus, $f(5)=\frac{1}{12}$.