Explanation

(d)
Given differential equation can be rewritten as
$\frac{ydy}{y + 1} = \frac{{e\ }^{x}dx}{e^{x} + 1}$
$\Rightarrow$ $\left( 1 - \frac{1}{y + 1} \right)dy = \frac{e^{x}}{e^{x} + 1}dx$
$\Rightarrow$ $y - \log{(y + 1)} = \log{(e^{x} + 1)} - \log c$ [integrating]
$\Rightarrow$ $y = \log\frac{\left( e^{x} + 1 \right)(y + 1)}{c}$
$\Rightarrow$ $\left( e^{x} + 1 \right)(y + 1) = ce^{y}$