If the solution curve, of the differential equation $$\frac{dy}{dx} = \frac{x + y - 2}{x - y}$$ passing through the point $$(2, 1)$$ is $$\tan^{-1}\frac{y-1}{x-1} - \frac{1}{\beta}\log_e\left(\alpha + \left(\frac{y-1}{x-1}\right)^2\right) = \log_e(x-1)$$, then $$5\beta + \alpha$$ is equal to

$$\frac{dy}{dx} = \frac{x+y-2}{x-y}$$. Substituting $$X = x-1, Y = y-1$$:

$$\frac{dY}{dX} = \frac{X+Y}{X-Y}$$

Let $$Y = vX$$: $$v + X\frac{dv}{dX} = \frac{1+v}{1-v}$$

$$X\frac{dv}{dX} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$$

$$\frac{1-v}{1+v^2}dv = \frac{dX}{X}$$

$$\int \frac{1}{1+v^2}dv - \int \frac{v}{1+v^2}dv = \ln|X| + C$$

$$\tan^{-1}v - \frac{1}{2}\ln(1+v^2) = \ln|X| + C$$

Substituting back ($$v = \frac{Y}{X} = \frac{y-1}{x-1}$$):

$$\tan^{-1}\frac{y-1}{x-1} - \frac{1}{2}\ln\left(1+\left(\frac{y-1}{x-1}\right)^2\right) = \ln|x-1| + C$$

Using $$(x,y) = (2,1)$$: $$\tan^{-1}(0) - \frac{1}{2}\ln(1) = \ln 1 + C$$, so $$C = 0$$.

Comparing with the given form: $$\beta = 2, \alpha = 1$$.

$$5\beta + \alpha = 10 + 1 = 11$$.

The answer is $$\boxed{11}$$.