Explanation

(c)
$\frac{dy}{dx} = \frac{x + y + 1}{x + y - 1}$
Put $x + y = t$
$\Rightarrow$ $1 + \frac{dy}{dx} = \frac{dt}{dx}$
$\Rightarrow$ $\frac{dy}{dx} = \frac{dt}{dx} - 1$
$\therefore$ $\frac{dt}{dx} - 1 = \frac{t + 1}{t - 1}$
$\Rightarrow$ $\frac{dt}{dx} = \frac{t + 1 + t - 1}{t - 1}$
$\Rightarrow$ $\frac{dt}{dx} = \frac{2t}{t - 1}$
$\Rightarrow$ $\left( \frac{t - 1}{2t} \right)dt = dx$
$\Rightarrow$ $\left( \frac{1}{2} - \frac{1}{2t} \right)dt = dx$
On integrating, we get
$\frac{1}{2}t - \frac{1}{2}\log t = x + c_{1}$
$\Rightarrow$ $t - \log t = 2x + 2c_{1}$
$\Rightarrow$ $x + y - \log{(x + y)} = 2x + 2c_{1}$
$\Rightarrow$ $y = x + \log{(x + y) + c}$