We have $y'=(1+x)\,y+xy^2$.

Rewriting this gives $y'-(1+x)\,y=xy^{2}$. This is a standard Bernoulli diffeq form $y'+a(x)\,y=b(x)\,y^{\alpha}$, which is solved by performing the substitution $\phi=y^{1-\alpha}$. Using this substitution, which for our specific case is $\phi = y^{1-2} = y^{-1}$, we have
$\displaystyle -\frac{y'}{y^2} +\frac{x+1}{y} = -x \Longleftrightarrow \phi' + (x+1)\,\phi = -x$.

To make this substitution more rigorous (since it is less obvious), differentials can be used. Also, to account for this division and ensure we miss no solutions, we make sure to check when the denominators are equal to 0, which is $y=0$. Since plugging in $y=0$ satisfies the diffeq, this is a constant solution. Then, using the integrating factor $\exp\left(\displaystyle\int x+1\text{ d}x\right) = e^{\frac{x^2}{2}+x}$, we get
$$
\begin{align*}
    \phi'e^{\frac{x^2}{2}+x} + \left(e^{\frac{x^2}{2}+x}\right)'\phi &= -xe^{\frac{x^2}{2}+x}\\
    \int\left(\phi \,e^{\frac{x^2}{2}+x}\right)'\text{d}x &= \int-xe^{\frac{x^2}{2}+x}\text{ d}x\\
    \phi \, e^{\frac{x^2}{2}+x} &= \sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C\\
    \phi &= \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}}
\end{align*}
$$
Undoing our substitution, we have, for the positive Dawson Integral $D_+(x) = \frac{\sqrt{\pi}}2e^{-x^2}\operatorname{erfi}(x)$ that
$\displaystyle \phi = y^{-1} = \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} = \sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}$
$\displaystyle\implies \boxed{y(x)=0, \qquad \frac{1}{\sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}}}$