If $\text{⨍ (x)=}\underset{{\mathrm{x}}^2}{\overset{{\mathrm{x}}^2+1}{\int }}{\text{ e}}^{-{\mathrm{t}}^2}\text{ dt}$ then the function $\text{⨍ (x)}$ decreases in

$$\begin{gathered} {\text{⨍ }}^1{{\text{(x)=e}}^{-({\mathrm{x}}^2+1)}}^2{\text{-e}}^{-{\mathrm{x}}^{4^{.2^{\mathrm{x}}}}}\text{.2x} \\ =2\mathrm{x}\left({{\mathrm{e}}^{-({\mathrm{x}}^2+1)}}^2-{\mathrm{e}}^{-{\mathrm{x}}^4}\right) \\ {\mathrm{x}}^{4+2{\mathrm{x}}^2}+1>{\mathrm{x}}^4 \\ {\left({\mathrm{x}}^2+1\right)}^2>{\mathrm{x}}^4 \\ -{\left({\mathrm{x}}^2+1\right)}^2<-{\mathrm{x}}^4 \end{gathered}$$

$\text{⨍ (x)}$ is decreases

$$\begin{gathered} \Rightarrow {\text{⨍ }}^1\text{(x)<0} \\ 2\mathrm{x}\left({\mathrm{e}}^{-{\left({\mathrm{x}}^2+1\right)}^2}-{\mathrm{e}}^{-{\mathrm{x}}^4}\right)<0 \\ 2\mathrm{x}>0 \Rightarrow \mathrm{x}>0 \end{gathered}$$