This is a differential equation of the form

$\dfrac {dy}{dx}+ Py=Q$

$\therefore P=\dfrac {4x}{x^2+1},Q=\dfrac 1{(x^2+1)^3}$

$I.F.=e^{\dfrac {4x}{x^2+1}}dx$

$=e^{2\log(x^2+1)^2}$

$=(x^2+1)^2$

The solution is,

$y\times(x^2+1)^2=\int\dfrac 1{(x^2+1)^3}\times(x^2+1)^2dx$

$=\int\dfrac 1{x^2+1}dx$

$=\tan^{-1}x+ c$

$\therefore y=\dfrac 1{(x^2+1)^2}\tan^{-1}x+\dfrac c{(x^2+1)^2}$

$\therefore y-\dfrac 1{(x^2+1)^2}\tan^{-1}x=\dfrac c{(x^2+1)^2}$