= Solution
For real solutions work on $x>-1$. Dividing by $2(1+x)$ puts the <linear ordinary differential equation> into the form
$$
y'+\left(-1+\frac1{2(1+x)}\right)y=\frac{x^2}{2\sqrt{1+x}}.
$$
Its <integrating factor> is
$$
\mu(x)=\exp\left(-x+\frac12\ln(1+x)\right)=e^{-x}\sqrt{1+x}.
$$
The <product rule> therefore gives
$$
(\mu y)'=\frac12x^2e^{-x}.
$$
Twice using <integration by parts>, or differentiating the proposed antiderivative, gives $\int x^2e^{-x}\,dx=-e^{-x}(x^2+2x+2)$. Hence
$$
\boxed{y(x)=\frac{Ce^x-\tfrac12(x^2+2x+2)}{\sqrt{1+x}},\qquad x>-1.}
$$
The endpoint $x=-1$ is singular. A finite continuous extension would require $C=e/2$, but then, with $h=x+1\downarrow0$, $y=\tfrac12\sqrt h+O(h^{3/2})$. Its derivative diverges, so even this choice has no differentiable extension to that endpoint.
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