jmc

We can write $$\begin{array}{rl}f(x)& =x+\frac{x}{x^2+1}+\frac{x(x+4)}{x^2+2}+\frac{2(x+2)}{x(x^2+2)}\\ & =\frac{x(x^2+1)+x}{x^2+1}+\frac{x^2(x+4)}{x(x^2+2)}+\frac{2(x+2)}{x(x^2+2)}\\ & =\frac{x^3+2x}{x^2+1}+\frac{x^3+4x^2+2x+4}{x(x^2+2)}\\ & =\frac{x(x^2+2)}{x^2+1}+\frac{4x^2+4}{x(x^2+2)}+\frac{x(x^2+2)}{x(x^2+2)}\\ & =\frac{x(x^2+2)}{x^2+1}+4\cdot \frac{x^2+1}{x(x^2+2)}+1.\end{array}$$By AM-GM, $$\frac{x(x^2+2)}{x^2+1}+4\cdot \frac{x^2+1}{x(x^2+2)}\ge 2\sqrt{\frac{x(x^2+2)}{x^2+1}\cdot 4\cdot \frac{x^2+1}{x(x^2+2)}}=4,$$so $f(x)\ge 5.$

Equality occurs when $$\frac{x(x^2+2)}{x^2+1}=2,$$or $x(x^2+2)=2x^2+2.$ This simplifies to $x^3-2x^2+2x-2=0.$

Let $g(x)=x^3-2x^2+2x-2.$ Since $g(1)=-1$ and $g(2)=2,$ there exists a root of $g(x)=0$ between 1 and 2. In particular, $g(x)=0$ has a positive root.

Therefore, the minimum value of $f(x)$ for $x>0$ is $\boxed{5}.$