The general Solution of differential equation
$\left( {{e^x} + 1} \right)ydy = (y + 1){e^x}dx$ is

Given differential equation

$\left( {{e^x} + 1} \right)ydy = (y + 1){e^x}dx$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{{e^x}(1 + y)}}{{\left( {{e^x} + 1} \right)y}} \Rightarrow \frac{{dx}}{{dy}} = \frac{{\left( {{e^x} + 1} \right)y}}{{{e^x}(1 + y)}}$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \frac{{{e^x}y}}{{{e^x}(1 + y)}} + \frac{y}{{{e^x}(1 + y)}}$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \frac{y}{{1 + y}} + \frac{y}{{(1 + y){e^x}}}$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \frac{y}{{1 + y}}\left( {1 + \frac{1}{{{e^x}}}} \right)$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \frac{y}{{1 + y}}\left( {\frac{{{e^x} + 1}}{{{e^x}}}} \right)$

$\Rightarrow$ $\left( {\frac{y}{{1 + y}}} \right)dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)dx$

On integrating both sides,

we get
$\int {\frac{y}{{1 + y}}} dy = \int {\frac{{{e^x}}}{{1 + {e^x}}}} dx$

$\Rightarrow$ $\int {\frac{{1 + y - 1}}{{1 + y}}} dy = \int {\frac{{{e^x}}}{{1 + {e^x}}}} dx$

$\Rightarrow$ $\int 1 dy - \int {\frac{1}{{1 + y}}} dy = \int {\frac{{{e^x}}}{{1 + {e^x}}}} dx$

$\Rightarrow$ $y - \log |(1 + y) = \log |\left( {1 + {e^x}} \right) + \log k$

$\Rightarrow$ $y = \log (1 + y) + \log \left( {1 + {e^x}} \right) + \log (k)$

$\Rightarrow$ $y = \log \left\{ {k(1 + y)\left( {1 + {e^x}} \right)} \right\}$