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 $c, d \in \mathbb{R}(x + \frac{1}{x})$?
 A: 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*}
A: This is just a complement to Sangchul Lee’s answer. 
Lemma: Let $L/K$ be a finite separable field extension and let $\Sigma$ denote the set of all $K$-invariant field homomorphisms $L\rightarrow \overline L$ into an algebraic closure $\overline L$ of $L$. Then for each $x\in L$ we have ${\rm N}(x):= \prod_{\sigma\in\Sigma}\sigma(x)\in K$. In particular, for $x\neq 0$ we have
$$
x^{-1} = \frac1{{\rm N}(x)}\cdot \prod_{\sigma\neq \rm id} \sigma(x).
$$
A proof of this can be found in (almost?) every book on algebraic number theory, e.g. [Jürgen Neukirch, Algebraic Number Theory; Proposition 2.6].
So let’s apply this to $L = \mathbb R(x)$ and $K = \mathbb R(x^n+\frac1{x^n})$. The minimal polynomial of $x$ over $\mathbb R(x^n + \frac1{x^n})$ is 
$$
P(y):= y^{2n} - y^n\cdot \left(x^n+\frac1{x^n}\right) +1\in \mathbb R\left(x^n+\frac1{x^n}\right)[y].
$$
If $\zeta$ denotes a primitive $n$-th root of unity, then all zeros of $P(y)$ are $\zeta^i\cdot x^{\pm 1}$, $i=0,\dotsc,n-1$. If we denote by $\tau,\sigma_i\colon \mathbb R(x)\rightarrow \overline{\mathbb R(x)}$ the homomorphisms given by $\sigma_i(x):= \zeta^ix$ resp. $\tau(x) := x^{-1}$, then we have
$$
\Sigma = \{\sigma_0,\dotsc,\sigma_{n-1}, \sigma_0\tau,\dotsc, \sigma_{n-1}\tau\}
$$
with the notations as above. Hence, given any $f(x)\in \mathbb R(x)$, its inverse can be written as
$$
\frac1{f(x)}=\frac1{{\rm N}(f(x))}\cdot \prod_{i=1}^{n-1}f(\zeta^ix)\cdot \prod_{i=0}^{n-1} f(\zeta^ix^{-1}).
$$
Some more fine tuning might be necessary here; I will look into it later. 
EDIT: I don't see how to simplify this calculation, though. What can be said is that $\prod_{i=0}^{n-1}f(\zeta^ix^{-1})$ is a polynomial in $x^{-n}$ (apply the Lemma above to $L = \mathbb R(x) = \mathbb R(x^{-1})$ and $K = \mathbb R(x^{-n})$; the minimal polynomial of $x^{-1}$ over $K$ is $y^n - x^{-n}$, whose roots are $\zeta^ix^{-1}$, $i=0,\dotsc,n-1$. Therefore, each $K$-invariant field homomorphism $L\rightarrow \overline L$ belongs to the one induced by $x^{-1}\mapsto \zeta^ix^{-1}$ for some $i=0,\dotsc,n-1$. The Lemma then tells us that ${\rm N}(f(x^{-1})) = \prod_{i=0}^{n-1}f(\zeta^ix^{-1})\in K = \mathbb R(x^{-n})$). But still calculating this by hand will be painful.
I guess, it’s more practicable to use the following standard procedure: Since $\mathbb R(x)$ is a $2n$-dimensional vector space over $\mathbb R(x^n+\frac 1{x^n})$, given any $f(x)\in \mathbb R(x)$, the sequence $1, f(x), f(x)^2, \dotsc, f(x)^{m}$ for $m=2n$ will be linearly dependent over $\mathbb R(x^n+\frac1{x^n})$. (Chances are high that it suffices for $m$ to be a proper divisor of $2n$.) Hence, by some linear algebra, there are $a_0,\dotsc,a_{m}\in \mathbb R(x^n+\frac1{x^n})$, $a_0\neq 0$, such that
$$
a_{m}(f(x))^m + \dotsb + a_1 f(x) + a_0 = 0.
$$
Then, $f(x)^{-1}$ is given by
$$
f(x)^{-1} = -a_0^{-1}\cdot \bigl(a_1 + a_2 f(x) + \dotsb + a_m f(x)^{m-1}\bigr).
$$
This has the advantage that the calculation happens in $\mathbb R(x)$, whereas in the first approach the inverse is calculated in $\mathbb C(x)$.
