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I made a bunch of problems exercising the Lambert W-function in the solution, because I like to exercise to new concepts that I learn about. One that I came up with was rearranging $y = x^x(\ln x + 1)$, i.e. the derivative of $x^x$, for $x$. Is it even possible to do so, or do all the $x$'s make it too complex for even $W$?

I tried approaching it like I did for $y = x \ln x - x$ (the integral of $\ln x$), which involved trying to factor something out, but I quickly found myself running in circles. My scraps of an attempt at $y = x^x(\ln x + 1)$ follow.

$\begin{align} y &= x^x(\ln x + 1) \\ &= e^{x \ln x}(\ln x + 1) \\ &= e^{x \ln x}e^{\ln(\ln x + 1)} \\ &= e^{x \ln x + \ln(\ln x + 1)} \\ \ln y &= x \ln x + \ln(\ln x + 1) \\\\ &\qquad\text{???} \end{align}$

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    I don't believe $x^x (\log x+1)$ has an elementary inverse, if that's what you mean.2012-10-02

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For applying only Lambert W and elementary functions, your equation should be in the form

$f_1(f_2(x)e^{f_2(x)})=c,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where $c$ constant and $f_1$ and $f_2$ are elementary functions with a suitable elementary local inverse. Unfortunately your equation doesn't seem to be able to be brought into this form.