The solution curve of the differential equation, $$(1 + e^{-x})(1 + y^2)\frac{dy}{dx} = y^2$$ which passes through the point (0, 1), is

We have the differential equation

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

First we separate the variables. Dividing both sides by $$y^2$$ and by $$(1+e^{-x})$$, we get

$$ \left(\frac{1 + y^2}{y^2}\right)dy = \frac{dx}{1 + e^{-x}}. $$

Now we simplify the factor on the left. Because $$1 + y^2$$ divided by $$y^2$$ equals $$1/y^2 + 1$$, we can write

$$ \left(\frac{1}{y^2} + 1\right)dy = \frac{dx}{1 + e^{-x}}. $$

Next we integrate both sides. We state the basic antiderivative formulas we will use:

$$\int y^{-2}\,dy = -y^{-1} + C, \qquad \int 1\,dy = y + C,$$

and for the $$x$$-integral we note that

$$\int \frac{dx}{1 + e^{-x}} = \int \frac{e^{x}\,dx}{e^{x} + 1}.$$

Using the substitution $$t = e^{x} + 1$$ (so $$dt = e^{x}dx$$), this integral becomes $$\int \frac{dt}{t} = \ln t + C = \ln(1 + e^{x}) + C.$$

Carrying out the integrations, we obtain

$$ -\,\frac{1}{y} + y = \ln(1 + e^{x}) + C. $$

The solution must pass through the point $$(0,1)$$. Substituting $$x = 0$$ and $$y = 1$$ gives

$$ -\,\frac{1}{1} + 1 = \ln(1 + e^{0}) + C \;\;\Longrightarrow\;\; 0 = \ln 2 + C. $$

Hence

$$ C = -\ln 2. $$

Putting this value of $$C$$ back into the integrated equation gives

$$ -\,\frac{1}{y} + y = \ln(1 + e^{x}) - \ln 2. $$

Using the logarithm law $$\ln a - \ln b = \ln\!\left(\dfrac{a}{b}\right)$$, we write

$$ -\,\frac{1}{y} + y = \ln\!\left(\frac{1 + e^{x}}{2}\right). $$

To eliminate the fraction on the left, we multiply the entire equation by $$y$$:

$$ -1 + y^2 = y \,\ln\!\left(\frac{1 + e^{x}}{2}\right). $$

Finally, adding $$1$$ to both sides, we arrive at

$$ y^2 = 1 + y \,\ln\!\left(\frac{1 + e^{x}}{2}\right). $$

This equation matches Option C.

Hence, the correct answer is Option C.