3 It is given that \(x = \ln ( 1 - y ) - \ln y\), where \(0 < y < 1\).
- Show that \(y = \frac { \mathrm { e } ^ { - x } } { 1 + \mathrm { e } ^ { - x } }\).
- Hence show that \(\int _ { 0 } ^ { 1 } y \mathrm {~d} x = \ln \left( \frac { 2 \mathrm { e } } { \mathrm { e } + 1 } \right)\).