5. If the positive number $x$ satisfies $x^{5}-x^{3}+x=2$, then the range of $x$ is ( ).
(A) $\sqrt[6]{3}\sqrt[3]{2}$
5.A.

Obviously, $x \neq 0, x^{6}+1=\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)=$ $\frac{x^{2}+1}{x}\left(x^{5}-x^{3}+x\right)=2\left(x+\frac{1}{x}\right)$.
Since $x>0$ and $x \neq 1$, we have
$$
x^{6}+1=2\left(x+\frac{1}{x}\right)>4, x>\sqrt[6]{3} \text {. }
$$

From the given condition, we get $\frac{2}{x^{3}}+1=x^{2}+\frac{1}{x^{2}}>2$, then $x<\sqrt[3]{2}$. Therefore, $\sqrt[6]{3}<x<\sqrt[3]{2}$.