Explanation

(c)
Given differential equation can be rewritten as
$\frac{dy}{dx} = \frac{x + y - 1}{x + y + 1}$
Put $x + y = t$
$\Rightarrow$ $\frac{dy}{dx} = \frac{dt}{dx} - 1$
$\therefore$ $\frac{dt}{dx} - 1 = \frac{t - 1}{t + 1}$
$\Rightarrow$ $\frac{1}{2}\ \left( t + \log t \right) = x + \frac{c}{2}$
$\Rightarrow$ $\frac{1}{2}\ \left( t + \log t \right) = x - y + c$
$\Rightarrow$ $\log{(x + y)} = x - y + c$