Let $f:R\to R$  and $g:R\to R$  be two functions defined by $f\left(x\right)={\log }_e(x^2+1)-e^{-x}+1$ and $g\left(x\right)=\frac{1-2e^{2x}}{e^x}$ . Then the number of integral values of  $\alpha$ for which of the following range of  $\alpha ,$  the inequality $f\left(g\left(\frac{{(\alpha -1)}^2}{3}\right)\right)>f\left(g(\alpha -\frac{5}{3})\right)$ holds is

Given function is  $f\left(x\right)={\log }_e(x^2+1)-e^{-x}+1$
Differentiate w.r.t $x$ both sides.
$\Rightarrow f\text{'}\left(x\right)=\frac{2x}{x^2+1}+e^{-x}>0,\forall x\in R$ 
For every real value of $f\left(x\right)$ is strictly increasing 
Given function $g\left(x\right)=\frac{1-2e^{2x}}{e^x}=e^{-x}-2e^x$ 
Differentiate w.r.t. $x$  both sides. 
$\Rightarrow g\text{'}\left(x\right)=-(2e^x+e^{-x})<0,\forall x\in R$ 
$\Rightarrow g$  is decreasing. 
Take    $f\left(g\left(\frac{{(\alpha -1)}^2}{3}\right)\right)>f\left(g(\alpha -\frac{5}{3})\right)$
$\Rightarrow g\left(\frac{{(\alpha -1)}^2}{3}\right)>g(\alpha -\frac{5}{3})$ 
 $\Rightarrow \frac{{(\alpha -1)}^2}{3}<\alpha -\frac{5}{3}$
 $\Rightarrow {\alpha }^2-5\alpha +6<0\Rightarrow (\alpha -2)(\alpha -3)<0\Rightarrow \alpha \in (2,3)$