Let the solution curve $$y = y(x)$$ of the differential equation $$(1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1$$ pass through the point $$\left(0, \frac{\pi}{2}\right)$$. Then, $$\lim_{x \to \infty} e^x y(x)$$ is equal to

We have the differential equation $$(1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1$$, which we rewrite as $$\frac{dy}{dx} + y = \frac{1}{1+e^{2x}}$$. This is a first-order linear ODE of the form $$y' + P(x)y = Q(x)$$ with $$P(x) = 1$$ and $$Q(x) = \frac{1}{1+e^{2x}}$$.

The integrating factor is $$\mu = e^{\int 1\,dx} = e^x$$. Multiplying both sides by $$e^x$$:

$$\frac{d}{dx}(e^x y) = \frac{e^x}{1+e^{2x}}$$

We integrate the right side. Substituting $$u = e^x$$, $$du = e^x\,dx$$:

$$\int\frac{e^x}{1+e^{2x}}\,dx = \int\frac{du}{1+u^2} = \tan^{-1}(u) = \tan^{-1}(e^x) + C$$

So the general solution is $$e^x y = \tan^{-1}(e^x) + C$$, giving $$y = e^{-x}\left[\tan^{-1}(e^x) + C\right]$$.

Now we apply the initial condition: the curve passes through $$\left(0, \frac{\pi}{2}\right)$$. At $$x = 0$$:

$$\frac{\pi}{2} = e^0\left[\tan^{-1}(e^0) + C\right] = \tan^{-1}(1) + C = \frac{\pi}{4} + C$$

So $$C = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$.

The particular solution is $$y = e^{-x}\left[\tan^{-1}(e^x) + \frac{\pi}{4}\right]$$, and therefore $$e^x y = \tan^{-1}(e^x) + \frac{\pi}{4}$$.

Now we compute $$\lim_{x \to \infty} e^x y(x) = \lim_{x \to \infty}\left[\tan^{-1}(e^x) + \frac{\pi}{4}\right]$$. As $$x \to \infty$$, $$e^x \to \infty$$, so $$\tan^{-1}(e^x) \to \frac{\pi}{2}$$.

Therefore $$\lim_{x \to \infty} e^x y(x) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$$.

Hence, the correct answer is Option B.