Answer
\[y(x)=\frac{4\tan^{-1}x}{1+x^2}+\frac{C}{1+x^2}\]
Work Step by Step
$\large\frac{dy}{dx}$+$\large\frac{2x}{1+x^2}$ $y=\large\frac{4}{(1+x^2)^2}$ ___(1)
(1) is Linear differential equation
Integrating Factor:-
\[I(x)=e^{\int\frac{2x}{1+x^2}dx}=e^{\ln|1+x^2|}=1+x^2\]
Multiply (1) by $I(x)$
\[(1+x^2)\frac{dy}{dx}+2xy=\frac{4}{1+x^2}\]
\[\frac{d}{dx}[y(1+x^2)]=\frac{4}{1+x^2}\]
Integrating,
\[y(1+x^2)=4\int \frac{dx}{1+x^2}+C\]
$C$ ia constant of integration
\[y(1+x^2)=4\tan^{-1}x+C\]
\[y(x)=\frac{4\tan^{-1}x}{1+x^2}+\frac{C}{1+x^2}\]
Hence General Solution of (1) is $y(x)\large=\frac{4\tan^{-1}x}{1+x^2}+\frac{C}{1+x^2}$
