$\lim_{n\to\infty}\frac{n^{3/2}}{\ln n}\int_0^1\frac{\sqrt x\ln x}{(1+2x-x^2)^n}dx$, an attempt to generalize Watson's lemma Background and Motivation
Given that Watson's lemma, a theorem that can compute the asymptotic expansion of $$\int_0^N p(x)e^{-zq(x)}dx\text{ as } z\to\infty,$$ with the elementary restriction of $p,q$ being $\color{red}{C^\infty}[0,\varepsilon)$ for some $\varepsilon>0$ (and some other restrictions), I want to generalize this theorem, to some extent.  

Question
  I've already evaluated the limit $$\lim_{n\to\infty}n^{3/2}\int_0^1\frac{\sqrt x}{(1+2x-x^2)^n}dx=\frac{\sqrt{2\pi}}8,$$ and I can even use Watson's lemma to calculate the asymptotic expansion to $O(n^\alpha)$ with arbitrary big $\alpha$.
  i) Given $\mu>0$, can we evaluate $$\lim_{n\to\infty}\frac{n^{3/2}}{\ln^\mu n}\int_0^1\frac{\sqrt x\ln^\mu (1/x)}{(1+2x-x^2)^n}dx?$$ Furthermore, is there a method to find all terms of the asymptotic expansion of the integral?
This integral is merely an example of the generalization of the theorem.
  ii) Reference request: Given $p(x)\in C^{\color{red}\omega}(0,\varepsilon)\cap C^{\color{red}\omega}(a-\varepsilon,a)\cap R[0,a]$, is there a method to find the full asymptotic expansion of $$\int_0^a p(x)e^{zx}dx$$ as $z\to\infty$?  

Attempt of (i)
Fix $\varepsilon>0$, noting that $\displaystyle\int_\varepsilon^1\frac{\sqrt x\ln ^\mu(1/x)}{(1+2x-x^2)^n}=O(n^{-N})$ for all $N\in\mathbb R$, this term can be neglected. There is a very classical method for this type of integrals which is substituting $\sqrt xdx=dt$, then obtaining $$\int_0^{\varepsilon'}\frac{\ln^\mu (1/x)}{(1+2x^{2/3}-x^{4/3})^n}dx$$ and finally neglect the higher order term. But this method can only be used when $\mu=0$ and can only get the rough order of it.
 A: Make the substitution $x = \phi(u) = 1 - \sqrt {2 - e^u}$. The integrand will become $f(u) e^{-n u}$. It can be proved that expanding $f$ around zero and integrating termwise will give a complete asymptotic series for large $n$. All the steps are the same as for the standard Laplace method.
For the leading term, we'll get
$$f(u) = \frac {e^u \sqrt {\phi(u)}} {2 (1 - \phi(u))} (-\ln \phi(u))^\mu \sim
\frac {\sqrt u} {2 \sqrt 2} (-\ln u)^\mu, \quad u \to 0^+, \\
I = \int_0^{\ln 2} f(u) e^{-n u} du \sim
\int_0^\infty \frac {\sqrt u} {2 \sqrt 2} (-\ln u)^\mu e^{-n u} du,
\quad n \to \infty.$$
For $\mu \in \mathbb N$, the last integral (times $2 \sqrt 2$) can be rewritten as
$$\int_0^\infty \sqrt u \, (-\ln u)^\mu e^{-n u} du =
\frac {d^\mu} {da^\mu} \int_0^\infty \sqrt u \,u^{-a} e^{-n u} du
 \bigg\rvert_{a = 0} = \\
\frac {d^\mu} {da^\mu} n^{a - 3/2} \Gamma(3/2 - a) \bigg\rvert_{a = 0} \sim
n^{a - 3/2} \Gamma(3/2 - a) \ln^\mu n \bigg\rvert_{a = 0},
\quad n \to \infty, \\
\lim_{n \to \infty} \frac {n^{3/2}} {\ln^\mu n} I =
\frac {\sqrt {2 \pi}} 8.$$
