Let f , g : 0 , ∞ → R be two functions defined by f x = ∫ − x x t − t 2 e − t 2 d t and g x = ∫ 0 x 2 t 1 2…

Given: $g\left(x\right)={\int }_0^{x^2}{t}^{\frac{1}{2}}{e}^{-t}dt$

Let, $t=y^2$

$\Rightarrow dt=2ydy$

$\Rightarrow g\left(x\right)={\int }_0^{x}2y^2{e}^{-y^2}dy$

$\Rightarrow g\left(x\right)=2{\int }_0^x{t}^2{e}^{-t^2}dt$

Also, $f\left(x\right)=2{\int }_0^x\left(t-t^2\right)e^{-t^2}dt$y

$\Rightarrow f\left(x\right)+g\left(x\right)=2{\int }_0^{x}t{e}^{-t^2}dt$

Let, $t^2=p$

$\Rightarrow 2tdt=dp$

$\Rightarrow f\left(x\right)+g\left(x\right)={\int }_0^{x^2}{e}^{-p}dp$

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

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

$\Rightarrow 9f\left(\sqrt{{\log }_{e}9}\right)+g\left(\sqrt{{\log }_{e}9}\right)=9\left(1-\frac{1}{9}\right)$

$\Rightarrow 9f\left(\sqrt{{\log }_{e}9}\right)+g\left(\sqrt{{\log }_{e}9}\right)=8$