(i) Start with the given equation: \(x = \ln(1-y) - \ln y\).

This can be rewritten using properties of logarithms: \(x = \ln \left( \frac{1-y}{y} \right)\).

Exponentiating both sides gives: \(e^x = \frac{1-y}{y}\).

Rearranging for \(y\), we get: \(y = \frac{1}{1 + e^x}\).

Substitute \(e^x = \frac{1}{e^{-x}}\) to obtain: \(y = \frac{e^{-x}}{1 + e^{-x}}\).

(ii) We need to evaluate \(\int_0^1 y \, dx\).

Using the expression for \(y\), we have \(y = \frac{e^{-x}}{1 + e^{-x}}\).

Let \(u = 1 + e^{-x}\), then \(du = -e^{-x} \, dx\), so \(dx = -\frac{du}{e^{-x}}\).

Substitute into the integral: \(\int y \, dx = \int \frac{e^{-x}}{u} \left(-\frac{du}{e^{-x}}\right) = -\int \frac{1}{u} \, du\).

This evaluates to \(-\ln |u| + C = -\ln(1 + e^{-x}) + C\).

Evaluate from 0 to 1:

At \(x = 1\), \(-\ln(1 + e^{-1})\).

At \(x = 0\), \(-\ln(1 + e^0) = -\ln(2)\).

Thus, \(\int_0^1 y \, dx = -\ln(1 + e^{-1}) + \ln(2)\).

Simplifying gives: \(\ln \left( \frac{2}{1 + e^{-1}} \right) = \ln \left( \frac{2e}{e+1} \right)\).