The solution of $\left(x^2+1\right)\frac{dy}{dx}+4xy=\frac{1}{x^2+1}$ is

 $$\begin{gathered} \left(x^2+1\right)\frac{dy}{dx}+4xy=\frac{1}{x^2+1} \\ \Rightarrow \frac{dy}{dx}+\frac{4x}{x^2+1}y=\frac{1}{{\left(x^2+1\right)}^2} \\ \Rightarrow \left(x^2+1\right)\frac{dy}{dx}+4xy=\frac{1}{x^2+1} \\ \Rightarrow \frac{dy}{dx}+\frac{4x}{x^2+1}y=\frac{1}{{\left(x^2+1\right)}^2} \end{gathered}$$

It is linear in y. Here $P=\frac{4x}{x^2+1},Q=\frac{1}{{\left(x^2+1\right)}^2}$

$\begin{array}{l}\int Pdx=\int \frac{4x}{x^2+1}dx=2\log \left(x^2+1\right)=\log {\left(x^2+1\right)}^2\\ \therefore \text{ I.F. }=e^{\int Pdx}=e^{\log {\left(x^2+1\right)}^2}={\left(x^2+1\right)}^2\end{array}$

$\therefore$ The solution is $y\left({\mathrm{IF}}_{.}\right)=\int Q\cdot \left(\mathrm{I}.\mathrm{F}.\right)dx$

$$\begin{gathered} \Rightarrow y{\left(x^2+1\right)}^2=\int \frac{1}{{\left(x^2+1\right)}^2}{\left(x^2+1\right)}^{2}dx=\int dx=x+c \\ \Rightarrow y{\left(x^2+1\right)}^2=x+c \end{gathered}$$