Solving a log equation: Is this equation analytic solvable? How can one solve this equation for $x$?  

$$1 = x (\ln x + \ln^2 x) $$

Can it be solved analytically? 
 A: My initial reaction was to think it could be solved using the Lambert W function. Unfortunately, it seems like one instead needs to use a generalization, the quadratic Lambert W function.
I'm using the notation from On the generalization of the Lambert W function by Mezo and Baricz. See also this question from MathOverflow, which is related.
Anyway, we want to solve:
$$
1=y\left[ \ln(y) + \ln(y)^2\right]
$$
First, transform (as suggested in the comments) via $y=\exp(x)$:
$$
1 = e^x(x+x^2)
$$
Now, the generalized Lambert W inverts
$$
f(x)=e^x(x-p)(x-q)
$$
And so it can be shown that the solution to
$$
c^2a=e^x(x-cp)(x-cq)
$$
is given by
$$
W_G\left( \begin{matrix} cp & cq \\ \, & \, \end{matrix} \,; \; c^2a  \right)
$$
where $W_G$ is the generalized Lambert W.
So for your case $1=e^x(x-0)(x-(-1))$, we can say:
$$
x = W_G\left( \begin{matrix} 0 & -1\\ \, & \, \end{matrix} \,; \; 1  \right)
$$
which seems like an exceptionally simple case (I'm certain it can be better simplified in fact) of the function. So it's an analytic solution in some sense :D
The linked paper, in addition to others, looks at many properties of the function, so maybe those would be more helpful for analysis
(and less likely than me to have made mistakes).

As an aside, one neat fact is that the Maclaurin series of $W_G$ is given by
$$
W_G\left( \begin{matrix} p & q \\ \, & \, \end{matrix} \,; \; a  \right)
=
p - \sum\limits_{n=1}^\infty \frac{1}{n!\,n}
\left[ \frac{an\exp(-p)}{q-p} \right]^n \;\mathcal{B}_{n-1}
\left( \frac{-2}{n(q-p)} \right)
$$
where $\mathcal{B}_{k}$ are the Bessel polynomials:
$$
\mathcal{B}_{n}(z)=\sum\limits_{k=0}^\infty \frac{(n+k)!}{k!(n-k)!}
\left( \frac{z}{2} \right)^k
$$
Thus (ignoring questions of convergence), we can write:
$$
x= \sum\limits_{n=1}^\infty \frac{-1}{n!\,n}
\left[ \frac{n}{-1} \right]^n \,
\sum\limits_{k=0}^\infty \frac{(n-1+k)!}{k!(n-1-k)!}
\left( \frac{1}{n} \right)^k
$$
Unfortunately, a quick check in Mathematica suggests it is outside the convergence radius.
