Given integral: $$\int \sqrt{\frac{x+7}{x}} \, dx$$

Let's make the substitution $$x = t^2$$. Then, $$dx = 2t \, dt$$.

Substituting these values, the integral becomes :

$$\int 2 \sqrt{t^2 + 7} \, dt$$

Now, let's evaluate this integral :

$$I(t) = 2\left(\frac{t}{2} \sqrt{t^2+7} + \frac{7}{2} \ln\left|t + \sqrt{t^2+7}\right|\right) + C$$

Substituting back $$t = \sqrt{x}$$, we have :

$$I(x) = 2\left(\frac{\sqrt{x}}{2} \sqrt{x+7} + \frac{7}{2} \ln\left|\sqrt{x} + \sqrt{x+7}\right|\right) + C$$

Simplifying further :

$$I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln\left|\sqrt{x} + \sqrt{x+7}\right| + C$$

We are given that $$I(9) = 12+7 \ln 7$$.

Let's substitute $$x = 9$$ and solve for the constant $$C$$:

$$12+7 \ln 7 = \sqrt{9} \sqrt{9+7}+7 \ln |\sqrt{9}+\sqrt{9+7}|+C$$

$$ \Rightarrow $$ $$12+7 \ln 7 = 3 \sqrt{16}+7 \ln |\sqrt{9}+\sqrt{16}|+C$$

$$ \Rightarrow $$ $$12+7 \ln 7 = 3 \cdot 4+7 \ln (3+\sqrt{16})+C$$

$$ \Rightarrow $$ $$12+7 \ln 7 = 12+7 \ln (3+4)+C$$

$$ \Rightarrow $$ $$12+7 \ln 7 = 12+7 \ln 7+C$$

From this equation, we can see that $$C = 0$$.

Now, we need to calculate $$I(1)$$ :

$$I(1) = \sqrt{1} \sqrt{1+7}+7 \ln |\sqrt{1}+\sqrt{1+7}|$$

$$ \Rightarrow $$ $$I(1) = 1 \sqrt{8}+7 \ln (1+\sqrt{8})$$

$$ \Rightarrow $$ $$I(1) = \sqrt{8}+7 \ln (1+2 \sqrt{2})$$

Therefore, $$\alpha = \sqrt{8}$$.

Finally, to find $$\alpha^4$$:

$$\alpha^4 = \left(\sqrt{8}\right)^4$$

$$ \Rightarrow $$ $$\alpha^4 = 8^2$$

$$ \Rightarrow $$ $$\alpha^4 = 64$$

Hence, $$\alpha^4$$ is equal to 64.