Find a basis for the space of quotients of polynomials over $\mathbb{R}$ 
$F$ is a field and define $F(t):=\lbrace \frac{p}{q} \mid p,q \in F[t], q\neq0 \rbrace$. Find a basis for $\mathbb{R}(t)$.

I know that the basis for the polynomials over $\mathbb{R}$ is $\lbrace 1,x,x^2,x^3,\ldots \rbrace$, so intuitively I guess the basis for $\mathbb{R}(t)$ is $\lbrace 1,x,x^2,\ldots,\frac{1}{x},\frac{1}{x^2},\ldots \rbrace$. Is this correct? If not, what is the right approach to find the basis?

Some elements of response.
The family of vectors $\mathcal F = \{1,t,t^2, \dots, \frac{1}{t}, \frac{1}{t^2}, \dots\}$ is linearly independent. However $\mathcal F$ does not span $\mathbb R(t)$. If that was the case, you would be able to find $a_0, \dots, a_n, b_{-1}, \dots, b_{-m}$ such that for all $t \in \mathbb R$
$$\frac{1}{1+t^2}=a_0 + a_1 t + \dots + a_n t^n + \frac{a_{-1}}{t} + \dots + \frac{a_{-m}}{t^m} \tag{1}$$ As $$\lim\limits_{t \to +\infty} \frac{1}{1+t^2} = \lim\limits_{t \to +\infty} \frac{a_{-1}}{t} = \dots = \lim\limits_{t \to +\infty} \frac{a_{-m}}{t^m} = 0$$ $(1)$ implies $a_0=a_1= \dots = a_n=0$. Hence you have for all $t \in \mathbb R$ $$\frac{1}{1+t^2}=\frac{a_{-1}}{t} + \dots + \frac{a_{-m}}{t^m} \tag{2}$$ You will verify that such equality cannot hold for all $t \in \mathbb R$. Hence $\frac{1}{1+t^2}$ doesn't belong to the linear span of $\mathcal F$.
To find a basis, the partial fraction decomposition should be of interest. In particular, consider following subsets of $\mathbb R(t)$ $$\begin{cases}
A &= \{1, t, t^2, \dots \}\\
B &= \left\{\frac{1}{(t-a)^n} \, | \, a \in \mathbb R, \, n \in \mathbb N \right\}\\
C &=\left\{\frac{1}{(t^2+2bt+c)^n}, \frac{t}{(t^2+2bt+c)^n} \, | \, b,c \in \mathbb R, \, b^2-c <0, \,n \in \mathbb N \right\}
\end{cases}$$ According to partial fraction decomposition, $A \cup B \cup C$ should be a basis of  $\mathbb R(t)$ as the irreducible polynomials of $\mathbb R[t]$ are precisely the $t-a$ and $t^2+2bt+c$ with $a,b,c \in \mathbb R$ and $b^2-c <0$.
