Answer
$$y = \frac{{\ln \left( {1 + {e^x}} \right)}}{{{e^x}}} + \frac{C}{{{e^x}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + y - \frac{1}{{1 + {e^x}}} = 0 \cr
& \frac{{dy}}{{dx}} + y = \frac{1}{{1 + {e^x}}} \cr
& \cr
& {\text{The equation is in the form }}\frac{{dy}}{{dp}} + p\left( x \right)y = q\left( x \right) \cr
& {\text{with }}p\left( x \right) = 1{\text{ and }}q\left( x \right) = \frac{1}{{1 + {e^x}}} \cr
& \cr
& {\text{Use the method of the integrating factors}}{\text{, calculate }}\mu = {e^{\int {p\left( x \right)dx} }} \cr
& \mu = {e^{\int {p\left( x \right)dx} }} = {e^{\int 1 dx}} \cr
& \mu = {e^x} \cr
& \cr
& {\text{Multiply both sides of the equation by }}\mu \cr
& {e^x}\frac{{dy}}{{dx}} + {e^x}y = \frac{{{e^x}}}{{1 + {e^x}}} \cr
& {\text{Express the result as }}\frac{d}{{dx}}\left[ {\mu y} \right] = {e^x}{\text{. Then}}{\text{,}} \cr
& \frac{d}{{dx}}\left[ {{e^x}y} \right] = \frac{{{e^x}}}{{1 + {e^x}}} \cr
& {\text{Integrate both sides}} \cr
& {e^x}y = \int {\frac{{{e^x}}}{{1 + {e^x}}}} dx \cr
& {e^x}y = \ln \left( {1 + {e^x}} \right) + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{\ln \left( {1 + {e^x}} \right)}}{{{e^x}}} + \frac{C}{{{e^x}}} \cr} $$
