As vertex is $V\left(1,\alpha \right)$, thus the quadratic is symmetric about $x=1$

$\Rightarrow f\left(1+x\right)=f\left(1-x\right)$
Replacing $\left(x\right)$ by $\left(x-1\right)$, we get,

$f\left(x\right)=f\left(2-x\right)$
Applying $\left(a+b-x\right)$ in $I$ and adding, we get,
$2I={\int }_0^2\frac{e^{f\left(x\right)}+e^{f\left(2-x\right)}}{e^{f\left(x\right)}+e^{f\left(2-x\right)}}dx$
$\Rightarrow 2I={\int }_0^{2}1dx={\left[x\right]}_0^2$
$\Rightarrow 2I=2-0$
$\Rightarrow I=1$