What is the inverse function of $e^x +x$? As the natural $\log(x)$ function is the inverse of the exponential $e^x$ and $\log(x +1)$ is the inverse of $e^x - 1$, what is the inverse of $e^x + x$?
 A: That inverse function cannot be expressed by the usual elementary functions, but Wolfram Alpha finds it to be
$$ f^{-1}(y) = y - W(e^y) $$
where $W$ is Lambert's W function.
This at least gives it a name, but may not otherwise seem like much progress because the W function is itself defined as the inverse of $xe^x$. So in some sense all we have achieved is relate the solution to your problem to the solution of another similarly difficult problem ...
A: 
To supplement the answer posted by @HenningMakholm, I thought it would be instructive to derive the inverse function in terms of Lambert's W Function.  To that end, we proceed.

First, we note that 
$$y=x+e^x\implies e^x=y-x \tag 1$$
Then, multiplying both sides of $(1)$ by $e^{y-x}$ reveals
$$e^y=(y-x)e^{y-x} \tag 2$$
Using the definition of Lambert's W Function,  $z=W(z)e^{W(z)}$, in $(2)$ indicates that $y-x=W(e^y)$, whereupon solving for $x$ yields
$$x=y-W(e^y)$$
Therefore, the inverse function of $f(x)=x+e^x$ is 
$$\bbox[5px,border:2px solid #C0A000]{f^{-1}(x)=x-W(e^x)}$$
which agrees with the result reported by @HenningMakholm!
A: Adding to the previous answers, let $y=f(x)=e^x+x,\;x\in\mathbb{R}^+$.
$$\begin{align*} f(x)&=y \\ e^x+x&=y \\ e^{e^x+x}&=e^y \\ e^xe^{e^x}&=e^y \\ e^x&=W(e^y) \\ x&=\log W(e^y) \\ f^{-1}(x)&=\log W(e^x)\end{align*}$$
One property of the Lambert W-function is that $\log W(z) =\log z-W(z)$ for positive $z$ in the principal real branch, which shows agreement with the other answers.
If you replace a constant with an arbitrary variable, you might be complicating things more than you think. $e^x+1$ is not difficult to invert, but $e^x+x$ (as we've seen) is a whole other story!
