Explanation

(d)
Given that, $\frac{dy}{dx} = 1 + x + y^{2} + xy^{2}$
This can be rewritten as, we get
$\frac{dy}{1 + y^{2}} = (1 + x)dx$
On integrating both sides, we get
$\int_{}^{}{\frac{dy}{1 + y^{2}} =}\int_{}^{}{(1 + x)dx}$
$\Rightarrow \tan^{- 1}{y = x + \frac{x^{2}}{2} + c}$
At $x = 0,\ y = 0$
$\Rightarrow 0 = 0 + 0 + c\ \Rightarrow c = 0$
$\therefore\tan^{- 1}{y = x + \frac{x^{2}}{2} \Rightarrow y = \tan\left( \ x + \frac{x^{2}}{2}\ \right)}$