Let $I\left(x\right)=\int \sqrt{\frac{x+7}{x}} dx$ and $I\left(9\right)=12+7{\log }_{e}7$. If $I\left(1\right)=\alpha +7{\log }_e\left(1+2\sqrt{2}\right),$ then ${\alpha }^4$ is equal to _____.

Given,

$I\left(x\right)=\int \sqrt{\frac{x+7}{x}} dx$

Now let, $\frac{x+7}{x}=t^2\Rightarrow -\frac{7}{x^2}dx=2t dt$

$\Rightarrow dx=\frac{-14t}{{\left(t^2-1\right)}^2}dt$

So, $I\left(x\right)=-14\int \frac{t^2}{{\left(t^2-1\right)}^2}dt$

$\Rightarrow I\left(x\right)=-14\int \frac{dt}{\left(t^2+{\displaystyle \frac{1}{t^2}}-2\right)}$

$\Rightarrow I\left(x\right)=\frac{-14}{2}\int \left[\frac{\left(1-{\displaystyle \frac{1}{t^2}}\right)}{{\left(t+{\displaystyle \frac{1}{t}}\right)}^2-4}+\frac{\left(1+{\displaystyle \frac{1}{t^2}}\right)}{{\left(t-{\displaystyle \frac{1}{t}}\right)}^2}\right]dt$

$\Rightarrow I\left(x\right)=-7\left(\frac{1}{4}\ln \left|\frac{t+{\displaystyle \frac{1}{t}}-2}{t+{\displaystyle \frac{1}{t}}+2}\right|-\frac{1}{t-{\displaystyle \frac{1}{t}}}\right)+c$

Now when $x=9, t=\frac{4}{3}$

$\Rightarrow I\left(9\right)= 12+7\times \ln 7=\frac{-7}{4}\ln {\left(\frac{1}{7}\right)}^2+7\times \frac{12}{7}+c$

$\Rightarrow c=\frac{7}{2}\ln 7$

Now when $x=1, t=2\sqrt{2}$

$\Rightarrow I\left(1\right)=+\frac{7}{4}\ln {\left(\frac{2\sqrt{2}+1}{2\sqrt{2}-1}\right)}^2+7\times \frac{2\sqrt{2}}{7}+\frac{7}{2}\ln 7$

$\Rightarrow I\left(1\right)=\frac{7}{2}\ln \left(\frac{{\left(2\sqrt{2}+1\right)}^2}{7}\right)+2\sqrt{2}+\frac{7}{2}\ln 7$

$\Rightarrow I\left(1\right)=7 \ln \left(2\sqrt{2}+1\right)-\frac{7}{2}\ln 7+2\sqrt{2}+\frac{7}{2}\ln 7$

$\Rightarrow \alpha =2\sqrt{2}\Rightarrow {\alpha }^4=64$