Expressing $f(x) = \frac{x-1}{x+1}$ as a sum of an even and odd function I am trying to write $$f(x) = \frac{1-x}{x+1}$$ as a sum of an even and odd function. One solution, though messy, is to use the following derivation:

We work backwards. Let $f = f_e + f_0$, where $f_e$ is even and $f_0$ odd. Then $f(x) = f_e (x) + f_0 (x)$ for all $x$ in the domain of $f$. But then
$$
f(-x) = f_e (-x) + f_0 (-x) = f_e (x) - f_0 (x).
$$
Adding, we eliminate $f_0 (x)$:
$$ 
f(x) + f(-x) = 2f_e (x),
$$
so
$$
f_e (x) = \frac{f(x) + f(-x)}{2}.
$$
Now, $f_0 = f(x) - f_e$. Then:
$$ 
f_0 (x) = f(x) - \frac{f(x) + f(-x)}{2} = \frac{2f(x) - f(x) - f(-x)}{2} = \frac{f(x) -f(-x)}{2}.
$$
One easily checks that $f_e$ is even and $f_0$ odd.

Using this derivation, I get:
\begin{align*}
\frac{1-x}{1+x} = \frac{1}{2}\left(\frac{1-x}{1+x} + \frac{1 + x}{1 - x} \right) + \frac{1}{2} \left(\frac{1-x}{1+x} - \frac{1+x}{1-x}  \right).
\end{align*}
Assuming I haven't made an algebra mistake, this seems to work, but it's not elegant. I did some research into this and found another proposed solution, but I believe it to be faulty. First, I'll just present it.
\begin{align*}
\frac{1-x}{1+x} & = \frac{1-x}{1+x} \cdot \frac{1-x}{1-x} \\
& = \frac{1 - 2x + x^2}{1 - x^2} \\
& = \frac{x^2 + 1}{1 - x^2} + \frac{2x}{1 - x^2}.
\end{align*}
One then checks that the first function is even and the second is odd.
My problem with this solution is that it is not valid for every $x$ in the domain of $f$. By definition, the domain of $f$ is $\mathbb{R} \setminus \{-1\}$. But this solution, in multiplying by $\frac{1-x}{1-x}$, presupposes that $x \neq 1$, which does not seem to me to be allowed.
I'm back at square one, then, because the solution I found does not seem in any way elegant or natural, and I'm assuming there is some kind of a trick that I am missing. I tried polynomial long division, partial fractions, and so forth, and nothing, other than the above derivation, brought me any progress.

In both cases, your solutions are undefined when $x=\pm1$. And that's natural. After all, $f(x)$ is undefined when $x=-1$ and, if you express $f(x)$ as $f_o(x)+f_e(x)$, with $f_o$ odd and $f_e$ even, then it is natural that at least one of the functions $f_o$ and $f_e$ is undefined when $x=-1$. But then, since $f_o(-x)=-f_o(x)$ and $f_e(-x)=f_e(x)$, if $f_o(x)$ is undefined when $x=-1$, then it is also undefined when $x=1$, and the same thing applies to $f_e(x)$.
