Approximation to the Lambert W function If:
$$x = y + \log(y) -a$$
Then the solution for $y$ using the Lambert W function is:
$$y(x) = W(e^{a+x})$$
In a paper I'm reading, I saw an approximation to this solution, due to "Borsch and Supan"(?):
$$y(x) = W(e^{a+x}) \approx x\left(1 - \frac{\log x - a}{1+x}\right)$$
Any idea how this approximation was derived?

First, let's consider $x=y+\log(y)$.
Assuming that $y$ is big, we can assume that $\log(y)$ is smaller than $y$. Therefore, we take a first estimate of $y=x$ and apply Newton's method to $f(y)=y+\log(y)-x$
$$
\begin{align}
y_{\text{next}}
&=y-\frac{f(y)}{f'(y)}\\
&=x-\frac{x+\log(x)-x}{1+\frac1x}\\
&=x\left(1-\frac{\log(x)}{x+1}\right)\tag{1}
\end{align}
$$
For $x\ge1$, this is a decent approximation for $\mathrm{W}(e^x)$.
Now consider that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{W}(e^x)=\frac{\mathrm{W}(e^x)}{\mathrm{W}(e^x)+1}\tag{2}
$$
Thus, for small $a$,
$$
\begin{align}
\mathrm{W}(e^{x{+}a})
&\approx\mathrm{W}(e^x)+a\frac{\mathrm{W}(e^x)}{\mathrm{W}(e^x)+1}\\
&=\mathrm{W}(e^x)\left(1+\frac{a}{\mathrm{W}(e^x)+1}\right)\\
&\approx\mathrm{W}(e^x)\left(1+\frac{a}{x+1}\right)\tag{3}
\end{align}
$$
Combining the approximations in $(1)$ and $(3)$, we get
$$
\mathrm{W}(e^{x{+}a})\approx x\left(1-\frac{\log(x)-a}{x+1}\right)\tag{4}
$$
