Given the equation: $ x^2 + \ln^2x = n $.
Let $x(n)$ - root of the equation and $x(n) > 1$
The problem: find missing terms in following sum: $$ x(n) \approx ... + O(\frac{\ln^4n}{n})$$
I've found the first term and get next result:
$$ x(n) \approx \sqrt{n} + ... + O(\frac{\ln^4n}{n}) $$
How to determine remaining terms of the sum?
Well, let $x=\sqrt{n}-y$, where $y=o(\sqrt n)$. Now: $$(\sqrt{n}-y)^2+\ln^2(\sqrt{n}-y)=n$$ $$-2\sqrt{n}\cdot y + y^2 + \ln^2\sqrt n + \left(\ln(1-\frac{y}{\sqrt n})\right)^2=0$$ Now discard the smaller terms and see what remains: $$-2\sqrt{n}\cdot y + \ln^2\sqrt n = 0$$ $$y = \left.\left({1\over2}\ln n\right)^2\right/2\sqrt{n}$$ (plus some small-o, of course).