Explanation

(a)
Given, $\frac{dy}{dx} = \frac{y - 1}{x^{2} + x}$
$\Rightarrow$ $\int_{}^{}\left( \frac{1}{x} - \frac{1}{x + 1} \right)dx = \int_{}^{}\frac{1}{y - 1}\ dy$
$\Rightarrow$ $\log x - \log{(x + 1)} = \log{(y - 1)} + \log c$
$\Rightarrow$ $\frac{x}{x + 1} = (y - 1)c$ …(i)
Since, this curve passes through $(1,0)c = - \frac{1}{2}$
$\therefore$ From Eq. (i) $2x + (y - 1)(x + !) = 0$