How to prove that $\int_0^\infty\frac{\left(x^2+x+\frac{1}{12}\right)e^{-x}}{\left(x^2+x+\frac{3}{4}\right)^3\sqrt{x}}\ dx=\frac{2\sqrt{\pi}}{9}$?

A Generalisation of the Integral:

Given a fixed $n\in\mathbb{N}$, let $A$, $P$, $Q$ be polynomials satisfying the following conditions:

1. $A(x)=Q(x)^{n+1}-2xP(x)Q(x)+P'(x)Q(x)-nP(x)Q'(x)$
2. $\deg A=\deg Q$
3. $P(0)=0, \ Q(0)\neq 0$
4. $A(x)Q(x)^{-(n+1)}=A(-x)Q(-x)^{-(n+1)}$

Then the integral of $e^{-x^2}A(x)Q(x)^{-(n+1)}$ over $\mathbb{R}^+$ can be computed as such. \begin{align} \int^\infty_0 e^{-x^2}\frac{A(x)}{Q(x)^{n+1}}\ dx &=\int^\infty_0 e^{-x^2}\left(1+\frac{-2xP(x)Q(x)+P'(x)Q(x)-nP(x)Q'(x)}{Q(x)^{n+1}}\right)\ dx\\ &=\frac{\sqrt{\pi}}{2}+\int^\infty_0\frac{((e^{-x^2})P(x))'Q(x)^n-e^{-x^2}P(x)(Q(x)^n)'}{Q(x)^{2n}}\ dx\\ &=\frac{\sqrt{\pi}}{2}+\left[e^{-x^2}\frac{P(x)}{Q(x)^n}\right]^\infty_0\\ &=\frac{\sqrt{\pi}}{2} \end{align} This means that as long as we can find polynomials $A$, $P$, $Q$ that satisfy all these conditions, we will be able to "construct" similar integrals to that which was posted in the question (at least in principle).

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Two Useful Facts:

Before we proceed to determine suitable $A$, $P$ and $Q$, we first prove the following facts:

$\text{Fact 1}$: $$\deg P=n\deg Q-1\tag{*}$$

To deduce this fact, observe that the polynomials $Q(x)^{n+1}$, $-2xP(x)Q(x)$, $P'(x)Q(x)$ and $-nP(x)Q'(x)$ have degrees $(n+1)\deg Q$, $\deg P+\deg Q+1$, $\deg P+\deg Q-1$ and $\deg P+\deg Q-1$ respectively. In order for the sum of these polynomials to have degree $\deg Q$, for all $\deg Q < j\leq\max((n+1)\deg Q, \deg P+\deg Q+1)$, the coefficients of the terms $x^j$ in each of these four polynomials have to add up to equal $0$. This requires $$\max((n+1)\deg Q, \deg P+\deg Q+1)=\min((n+1)\deg Q, \deg P+\deg Q+1)$$ and the desired result follows.

$\text{Fact 2}$: $$P(x)Q(x)^{-n}=-P(-x)Q(-x)^{-n}\tag{**}$$

This follows from Condition 4. Since $A(x)Q(x)^{-(n+1)}$ is even, $A(x)Q(x)^{-(n+1)}-1$ must also be even. But $A(x)Q(x)^{-(n+1)}-1=e^{x^2}(e^{-x^2}P(x)Q(x)^{-n})'$, so $P(x)Q(x)^{-n}$ must be odd.

Note that $(^{**})$ also implies Condition 3.

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A Simple Example: $n=1$, $\deg Q=2$

Let $n=1$ and $Q(x)=x^2+c\ $ with $c\neq 0$. By $(^{*})$ and $(^{**})$, $P$ is an odd polynomial of degree $1$, i.e. $P(x)=kx$ for some constant $k$. By Condition 1,

\begin{align} A(x) &=(x^2+c)^2-2kx^2(x^2+c)+k(x^2+c)-2kx^2\\ &=(1-2k)x^4+(2c(1-k)-k)x^2+c(k+c) \end{align} Since $\deg A=2$, $k=\frac{1}{2}$ and $c\neq\frac{1}{2}$. Thus $$A(x)=\left(c-\frac{1}{2}\right)x^2+c\left(c+\frac{1}{2}\right)$$ and we obtain the identity, for $c\in\mathbb{R}^+\setminus\{\frac{1}{2}\}$ $$\int^\infty_0e^{-x^2}\cdot\frac{x^2+\frac{c(2c+1)}{2c-1}}{(x^2+c)^2}\ dx=\frac{\sqrt{\pi}}{2c-1}$$

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The Case in Question: $n=2$, $\deg Q=4$

We follow the exact same procedure outlined above. In this case $n=2$ and $Q(x)=x^4+px^2+q$. Then $P(x)=rx^7+sx^5+tx^3+ux$. Applying Condition 1 and noting that $\deg A=4$ (i.e. the coefficients of $x^{12}$, $x^{10}$, $x^8$, $x^6$ are all $0$) ,

\begin{align} A(x) &=(x^4+px^2+q)^3-2x(rx^7+sx^5+tx^3+ux)(x^4+px^2+q)\\ &\ \ \ \ \ +(7rx^6+5sx^4+3tx^2+u)(x^4+px^2+q)-2(rx^7+sx^5+tx^3+ux)(4x^3+2px)\\ &=(p(3pq-t-2u)+q(3q+5s-2t)-7u)x^4+(3p(q^2-u)+q(3t-2u))x^2+q(q^2+u) \end{align} where \begin{align} &\ \ \ \ \ 1-2r=3p-2s-r(1+2p)=p(3p-2s)+r(3p-2q)+3q-3s-2t\\ &=p(p^2+6q+s-2t)+q(7r-2s)-5t-2u=0 \end{align} After some algebra, we may express $r,s,t,u$ in terms of the free variables $p,q$. $$(r,s,t,u)=\left(\frac{1}{2},\frac{4p-1}{4},\frac{4p^2-4p+8q+3}{8},\frac{-4p^2+16pq+12p-8q-15}{16}\right)$$ Therefore \begin{align} A(x) &=\tfrac{12p^2+16q^2-16pq-60p+24q+105}{16}x^4+\tfrac{12p^3-16p^2q+16pq^2-36p^2+64q^2-24pq+45p+48q}{16}x^2\\ &\ \ \ \ \ +\tfrac{16q^3-4p^2q+16pq^2-8q^2+12pq-15q}{16} \end{align} This yields, for $p,q,s,t,u\neq 0$, $$\small{\int^\infty_0e^{-x^2}\tiny{\frac{(12p^2+16q^2-16pq-60p+24q+105)x^4+(12p^3-16p^2q+16pq^2-36p^2+64q^2-24pq+45p+48q)x^2+(16q^3-4p^2q+16pq^2-8q^2+12pq-15q)}{16(x^4+px^2+q)^3}}\ dx=\frac{\sqrt{\pi}}{2}}$$ If $p=1$ and $q=\frac34$, this integral reduces to the one posted in the question.

The integrand has a closed-form anti-derivative: $${\large\int}\frac{\left(x^2+x+\frac1{12}\right)e^{-x}}{\left(x^2+x+\frac34\right)^3\sqrt x}\ dx=\frac{4\ e^{-x} \sqrt x}9\cdot\frac{8x^3+12x^2+18x-1}{\left(4x^2+4x+3\right)^2}+\frac{2\sqrt\pi}9 \,\operatorname{erf}\big(\sqrt{x}\big)\color{gray}{+C}$$ where $\operatorname{erf}(x)$ is the error function. This can be easily proved by differentiation. Using a well-known fact that $$\lim_{x\to\infty}\operatorname{erf}(x)=1$$ we get the result $$\int_0^\infty\frac{\left(x^2+x+\frac{1}{12}\right)e^{-x}}{\left(x^2+x+\frac{3}{4}\right)^3\sqrt{x}}\ dx=\frac{2\sqrt{\pi}}{9}.$$