Answer
$$y=\frac{c}{\sqrt{x^2+1}} $$
Work Step by Step
Given $$(x^2+1)\frac{dy}{dx} +xy=0$$
Rewriting the equation
$$ \frac{dy}{dx} +\frac{x}{x^2+1}y=0$$
Since
\begin{align*}
\mu(x)&=e^{\int \frac{x}{x^2+1}dx}\\
&=e^{\frac{1}{2}\ln(x^2+1) }\\
&=e^{\ln \sqrt{x^2+1}}\\
&= \sqrt{x^2+1}
\end{align*}
Then
\begin{align*}
y\mu(x)&=\int \mu(x) q(x)dx\\
\sqrt{x^2+1}y&=\int 0 dx\\
&=c
\end{align*}
Hence $$y=\frac{c}{\sqrt{x^2+1}} $$
