JEE Math Practice Question

The solution curve of the differential equation, (1 + e-x)(1 + y2)$${{dy} \over {dx}}$$ = y2, which passes through the point (0, 1), is :

1. A.y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$
2. B.y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$
3. C.y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$
4. D.y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$