Answer
$${y^2} + 2{e^y} + \ln \left( {{x^2} + 1} \right) = C$$
Work Step by Step
$$\eqalign{
& xdx + \left( {y + {e^y}} \right)\left( {{x^2} + 1} \right)dy = 0 \cr
& \left( {y + {e^y}} \right)\left( {{x^2} + 1} \right)dy = - xdx \cr
& {\text{Separate the variables}} \cr
& \left( {y + {e^y}} \right)dy = - \frac{x}{{{x^2} + 1}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {\left( {y + {e^y}} \right)dy} = - \int {\frac{x}{{{x^2} + 1}}} dx \cr
& \frac{1}{2}{y^2} + {e^y} = - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + {C_1} \cr
& {y^2} + 2{e^y} = - \ln \left( {{x^2} + 1} \right) + 2{C_1} \cr
& {y^2} + 2{e^y} + \ln \left( {{x^2} + 1} \right) = C \cr} $$
