Erf?

$\displaystyle \begin{aligned} I & = \int \frac{e^{-2x}}{\sqrt{x+5}} \ dx & \small \color{#3D99F6} \text{Let } u = \sqrt{x + 5} \implies dx = 2\sqrt{x+5} \ du \\ & = 2e^{10} \int e^{-2u^2} \ du & \small \color{#3D99F6} \text{Let } v = u\sqrt{2} \implies du = \frac{1}{\sqrt{2}} \ dv \\ & = \frac{2e^{10}}{2^{\frac 32}} \int 2e^{-v^2} \ dv \\ & = \frac{e^{10}\sqrt{\pi}}{\sqrt{2}} \int \frac{2e^{-v^2}}{\sqrt{\pi}} \ dv & \small \color{#3D99F6} \text{where} \int \frac{2e^{-v^2}}{\sqrt{\pi}} \ dv = \text{erf(}v\text{)} \\ & = \lim_{b \to \infty} \left. \sqrt{\frac{\pi}{2}}e^{10} \text{erf}(v) \ \right |_0^b \\ & = \lim_{b \to \infty} \left. \sqrt{\frac{\pi}{2}}e^{10}\text{erf}\left(u\sqrt{2}\right) \ \right|_0^b \\ & = \lim_{b \to \infty} \left. \sqrt{\frac{\pi}{2}}e^{10}\text{erf}\left(\sqrt{2}\sqrt{x+5}\right) \ \right|_0^b & \small \color{#3D99F6} \text{erf}\left(\infty\right) = 1 \\ & = \sqrt{\frac{\pi}{2}}e^{10}\left(1 - \text{ erf}\left(\sqrt{10}\right)\right) \end{aligned}$

$\implies a = 2, b = 5 \implies \dfrac{a + b + 2ab}{3} = \boxed{9}$

$\displaystyle I ~ =~ \int \frac{e^{-2x}}{\sqrt{x+5}} \ dx\\ \color{#3D99F6} {Let ~t^2 = 2*{x + 5}}~~ ~~~~~\implies~~~2x=t^2-10.~~~~~~~~\sqrt{x+5}=\dfrac t {\sqrt2},~~~and~~~dx=t.dt \\ \displaystyle \therefore~~I ~= ~\int \frac{\sqrt2*e^{-t^2}*e^{10} } t t.dt\\ \displaystyle~I~ = \sqrt2*e^{10} \int e^{-t^2}dt ~~~~~~\color{#3D99F6} {But~~erf(t)~=~ \int \dfrac {2* e^{-t^2}}{\sqrt{\pi}}dt }\\$ $\displaystyle \therefore~~I ~= ~ e^{10}* \sqrt{\dfrac {\pi} 2}* erf(t).\\ erf(t) |_0 ^\infty = \lim_{b \to \infty} erf(t) |_0 ^b~=1-erf(\sqrt{2* 5}).~~~~~~~~~~~~\color{#3D99F6} {\because~erf(\infty)=1.~~~~~t=\sqrt{2*(x+5)}}\\ \displaystyle \therefore ~~\int_0^{\infty} \frac{e^{-2x}}{\sqrt{x+5}} \ dx\\ = e^{2*5}*\sqrt{\dfrac {\pi} 2}*( 1-erf(\sqrt{2* 5}))\\ = e^{a*b}*\sqrt{\dfrac {\pi} a} *( 1-erf(\sqrt{a* b}) ) \\ \therefore~\dfrac{a+b+2*a*b} 3~=~\dfrac{2+5+2*2*5} 3 =\Large \color{#D61F06}{9}$