Let f, g: (0, ∞) → R be two functions defined by f(x) = ∫ _ - x ^ x...

Given: $g\left(x\right) = {∫}_{0}^{{x}^{2}}t\dfrac{1}{2}{e}^{- t}dt$

Let, $t = {y}^{2}$

$⇒ dt = 2ydy$

$⇒ g\left(x\right) = {∫}_{0}^{x}2{y}^{2}{e}^{- {y}^{2}}dy$

$⇒ g\left(x\right) = 2{∫}_{0}^{x}{t}^{2}{e}^{- {t}^{2}}dt$

Also, $f\left(x\right) = 2{∫}_{0}^{x}\left(t - {t}^{2}\right){e}^{- {t}^{2}}dt$y

$⇒ f\left(x\right) + g\left(x\right) = 2{∫}_{0}^{x}t{e}^{- {t}^{2}}dt$

Let, ${t}^{2} = p$

$⇒ 2tdt = dp$

$⇒ f\left(x\right) + g\left(x\right) = {∫}_{0}^{{x}^{2}}{e}^{- p}dp$

$⇒ f\left(x\right) + g\left(x\right) = 1 - {e}^{- {x}^{2}}$

$⇒ 9f\left(\sqrt{\log_{e}9}\right) + g\left(\sqrt{\log_{e}9}\right) = 9\left(1 - {e}^{- \log 9}\right)$

$⇒ 9f\left(\sqrt{\log_{e}9}\right) + g\left(\sqrt{\log_{e}9}\right) = 9\left(1 - \dfrac{1}{9}\right)$

$⇒ 9f\left(\sqrt{\log_{e}9}\right) + g\left(\sqrt{\log_{e}9}\right) = 8$