Since the degree of the dependent variable x is one it has to be reduced to the form

$ \frac{dx}{dy}+Px=Q$

Dividing throughout by $(1+y^2)$ we get,

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\tan^{-1}y}}{1+y^2} $

$ P=\frac{1}{(1+y^2)}$ and $Q=\frac{e^{\tan^{-1}y}}{1+y^2}$

$I.F=e^{\int P dy}$

$= e^{\int \frac{1}{(1+y^2)}dy}$

$ =e^{\tan^{-1}y}$

The solution is

$xe^{\tan^{-1}y}=\int \frac{e^{\tan^{-1}y}}{(1+y^2)}e^{\tan^{-1}y}dy+c $

For integrating on R.H.S

$Put\ \tan^{-1}y=t$

$\frac{1}{(1+y^2)}dy=dt$

$xe^{\tan^{-1}y}=\int e^{2t}dt+c$

$xe^{\tan^{-1}y}=\frac{1}{2} e^{2\tan^{-1}y}dt+c $