Field extension $\mathbb{R}(x) \supseteq \mathbb{R}(x + \frac{1}{x})$ 
What is a basis of $\mathbb{R}(x)$ over $ \mathbb{R}(x + \frac{1}{x})$?

It seems that $\mathbb{R}(x) = \{a + bx: a, b \in \mathbb{R}(x + \frac{1}{x}) \}$, but I cannot find the inverse of $a + bx$.
Is there any theorem that states, that we can find the inverse of $a + bx$ in the form $c + dx$, where $c, d \in \mathbb{R}(x + \frac{1}{x})$?

How about this?
\begin{align*}
\frac{1}{a+bx}
&= \frac{a + b(x+\frac{1}{x}) - bx}{a^2 + b^2 + ab(x + \frac{1}{x})} \\
&= \frac{a + b(x+\frac{1}{x})}{a^2 + b^2 + ab(x + \frac{1}{x})} \cdot 1 + \frac{-b}{a^2 + b^2 + ab(x + \frac{1}{x})} \cdot x
\end{align*}
