Integral $\int^{\infty}_0 \exp\left[-\left(4x+\frac{9}{x}\right)\right] \sqrt{x}\,dx$ 
How do I evaluate $$\displaystyle\int^{\infty}_0 \exp\left[-\left(4x+\dfrac{9}{x}\right)\right] \sqrt{x}\;dx?$$ 

To my knowledge the following integral should be related to the Gamma function.
I have tried using the substitution $t^2 = x$, and I got 
$$
2e^{12}\displaystyle \int^{\infty}_0 \exp\left[-\left(2t + \dfrac{3}{t}\right)^2\right] t^2 \; dt
$$ 
after substitution. But it seems like I can do nothing about this integral anymore. Can anyone kindly give me a hint, or guide me to the answer?

It looks like a tricky integral, however Feynman's trick deals with it nicely. 
$$I=\int^{\infty}_0 \exp\left(-\left(4x+\dfrac{9}{x}\right)\right) \sqrt{x}dx\overset{\sqrt x\to x}=2\int_0^\infty \exp\left(-\left(4x^2+\frac{9}{x^2}\right)\right)x^2 dx$$
Now consider the following integral:
$$I(t)=2\int_0^\infty \exp\left(-\left(4x^2+\frac{t}{x^2}\right)\right)x^2 dx$$
The reason why I'm putting the parameter in that place is because if $x^2$ is simplified then the integral becomes much easier. So let's take a derivative with respect to $t$ in order to get:
$$ I'(t)=-2\int_0^\infty \exp\left(-\left(4x^2+\frac{t}{x^2}\right)\right) dx=-\frac{\sqrt \pi}{2}e^{-4\sqrt t}$$
The above result follows using the Cauchy-Schlomilch transformation (see $3.3$).
I think that you are on the right track now and basically the future the steps would be to see that:
$$I(0)=\frac{\sqrt \pi}{16}\Rightarrow I=I(9)-I(0)+\frac{\sqrt\pi}{16}=-\frac{\sqrt \pi}2 \int_0^9e^{-4 \sqrt t}dt+\frac{\sqrt{\pi}}{16}=\boxed{\frac{13\sqrt \pi}{16e^2}}$$
