If the solution curve $$y = y(x)$$ of the differential equation $$(1 + y^2)(1 + \log_e x)dx + xdy = 0$$, $$x > 0$$ passes through the point $$(1, 1)$$ and $$y(e) = \frac{\alpha - \tan(\frac{3}{2})}{\beta + \tan(\frac{3}{2})}$$, then $$\alpha + 2\beta$$ is _______

We need to solve the differential equation $$(1+y^2)(1+\ln x)dx + xdy = 0$$ with $$y(1) = 1$$, then find $$\alpha + 2\beta$$.

Separate the variables.

Rearranging:

$$xdy = -(1+y^2)(1+\ln x)dx$$

$$\frac{dy}{1+y^2} = -\frac{(1+\ln x)}{x}dx$$

Integrate both sides.

Left side: $$\int \frac{dy}{1+y^2} = \tan^{-1}y$$

Right side: $$-\int \frac{1+\ln x}{x}dx$$. Let $$u = \ln x$$, $$du = dx/x$$:

$$-\int (1+u)du = -(u + u^2/2) = -\ln x - \frac{(\ln x)^2}{2}$$

So: $$\tan^{-1}y = -\ln x - \frac{(\ln x)^2}{2} + C$$

Apply initial condition $$y(1) = 1$$.

At $$x = 1$$: $$\tan^{-1}(1) = 0 - 0 + C$$, so $$C = \pi/4$$.

Find $$y(e)$$.

At $$x = e$$: $$\ln e = 1$$, so:

$$\tan^{-1}y = -1 - \frac{1}{2} + \frac{\pi}{4} = \frac{\pi}{4} - \frac{3}{2}$$

$$y = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right) = \frac{\tan(\pi/4) - \tan(3/2)}{1 + \tan(\pi/4)\tan(3/2)} = \frac{1 - \tan(3/2)}{1 + \tan(3/2)}$$

Comparing with $$y(e) = \frac{\alpha - \tan(3/2)}{\beta + \tan(3/2)}$$:

$$\alpha = 1$$, $$\beta = 1$$

$$\alpha + 2\beta = 1 + 2(1) = 3$$

The answer is 3.