An integrating factor

Rewrite. Divide by $(1+x^2)dx$:
$$\frac{dy}{dx} + \frac{x}{1+x^2}y = -x.$$

Integrating factor. Here $P(x) = \frac{x}{1+x^2}$, so
$$\int P(x)\,dx = \frac12 \ln(1+x^2) = \ln\sqrt{1+x^2}.$$
The integrating factor is
$$\mu(x) = e^{\int P\,dx} = \sqrt{1+x^2}.$$

Multiply.
$$\sqrt{1+x^2}\,y' + \frac{x}{\sqrt{1+x^2}}y = -x\sqrt{1+x^2}.$$
The left side is $\frac{d}{dx}(\sqrt{1+x^2}\,y)$.

Integrate.
$$\sqrt{1+x^2}\,y = -\int x\sqrt{1+x^2}\,dx = -\frac13(1+x^2)^{3/2} + C.$$
Hence
$$y(x) = -\frac13(1+x^2) + \frac{C}{\sqrt{1+x^2}}.$$