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

$$\begin{gathered} \text{I(x)=}\text{∫}\sqrt{\frac{\mathrm{x}+7}{\mathrm{x}}}\mathrm{dx} \\ \mathrm{Put} \mathrm{x}=7{\tan }^2\mathrm{\theta } \mathrm{dx}=14{\mathrm{tan\theta sec}}^2\mathrm{\theta d\theta } \\ \therefore \mathrm{I}=\int \sqrt{\frac{7(1+{\tan }^2\mathrm{\theta })}{7{\tan }^2\mathrm{\theta }}}14{\mathrm{tan\theta sec}}^2\mathrm{\theta }\hspace{0.17em}\mathrm{d\theta } \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}I=14\int {\sec }^3\mathrm{\theta }\hspace{0.17em}\hspace{0.17em}\mathrm{d\theta } \end{gathered}$$
Using the reduction formula to evaluate $\int {\sec }^3\mathrm{\theta }\mathrm{d\theta }$ , we get
$\mathrm{I}=7\left(\frac{\sqrt{\mathrm{x}(7+\mathrm{x})}}{7}+\ln \left(\frac{\sqrt{7+\mathrm{x}}}{\sqrt{7}}+\frac{\sqrt{\mathrm{x}}}{\sqrt{7}}\right)\right)$
Given, $\mathrm{I}\left(9\right)=12+7{\log }_{\mathrm{e}}7$
$\therefore 12+7{\log }_{\mathrm{e}}7=\frac{7\sqrt{9\left(16\right)}}{7}+\frac{7}{2}{\log }_{\mathrm{e}}\left(7\right)+\mathrm{c}$
$$\begin{gathered} \Rightarrow \mathrm{c}=\frac{7}{2}{\log }_{\mathrm{e}}7=7{\log }_{\mathrm{e}}\sqrt{7} \\ \mathrm{I}=7\left(\frac{\sqrt{\mathrm{x}(7+\mathrm{x})}}{7}+\ln \left(\frac{\sqrt{7+\mathrm{x}}}{\sqrt{7}}+\frac{\sqrt{\mathrm{x}}}{\sqrt{7}}\right)\right)+7{\log }_{\mathrm{e}}\sqrt{7} \\ \therefore \mathrm{I}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}(7+\mathrm{x})}+7{\log }_{\mathrm{e}}(\sqrt{\mathrm{x}+7}+\sqrt{\mathrm{x}}) \\ \mathrm{I}\left(1\right)=2\sqrt{2}+7{\log }_{\mathrm{e}}(2\sqrt{2}+1) \\ \therefore \mathrm{\alpha }=2\sqrt{2};{\mathrm{\alpha }}^4=64 \end{gathered}$$