<s>[INST] 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 friend gave me this integral as a challenge
$$
\int_0^\infty\frac{\left(x^2+x+\frac{1}{12}\right)e^{-x}}{\left(x^2+x+\frac{3}{4}\right)^3\sqrt{x}}\ dx=\frac{2\sqrt{\pi}}{9}.
$$
This integral can be written in an equivalent form 
$$
\int_0^\infty\frac{x^4+x^2+\frac{1}{12}}{\left(x^4+x^2+\frac{3}{4}\right)^3}e^{-x^2}\ dx=\frac{\sqrt{\pi}}{9}.
$$
I don't know how to prove this. I checked it numerically and it appears to be correct with 1000 digit accuracy. 
I tried several approaches. It seems this integral could be tackled by contour integration but so far I was unable to find a suitable contour. I also tried substitution, however with no luck.
Does anybody know how to calculate this integral?
 [/INST] 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
$$\int_0^\infty\frac{\left(x^2+x+\frac{1}{12}\right)e^{-x}}{\left(x^2+x+\frac{3}{4}\right)^3\sqrt{x}}\ dx=\frac{2\sqrt{\pi}}{9}.$$
</s>