Explanation

(d)
Given, $\frac{dy}{dx} = (4x + y + 1)^{2}$
Put $4x + y + 1 = v$
$\Rightarrow$ $\frac{dy}{dx} = \frac{dv}{dx} - 4$
$\therefore$ $\frac{dv}{dx} - 4 = v^{2}$
$\Rightarrow$ $\frac{dv}{v^{2} + 4} = dx$
$\Rightarrow$ $\frac{1}{2}\tan^{- 1}\left( \frac{v}{2} \right) = x + c$ [integrating]
$\Rightarrow$ $\tan^{- 1}\left( \frac{4x + y + 1}{2} \right) = 2x + c$
$\Rightarrow$ $4x + y + 1 = 2\tan{(2x + c)}$