Answer
$y= \frac{C+\arctan x}{1+x^2}$
Work Step by Step
We are given
$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{(1+x^2)^2}$
We can rewrite:
$(1+x^2)^2dy+(1+x^2)2xydx=dx$
$(2xy+2x^3y-1)dx+(1+x^2)^2dy=0$
Here, $M(x,y)=2xy+2x^3y-1$
$N(x,y)=(1+x^2)^2$
We have:
$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{(2x+2x^3)-4x(1+x^2)}{(1+x^2)^2}=\frac{-2x}{1+x^2}$
The integrating factor is:
$I(x)=e^{\int f(x)dx}=e^{\int \frac{-2x}{1+x^2} dx}=\frac{1}{1+x^2}$
Multiplying equation (1) by $I(x)=\frac{1}{1+x^2}$
$\frac{1}{1+x^2}[(2xy+2x^3y-1)dx+(1+x^2)^2dy]=0$
$(2xy-\frac{1}{1+x^2})dx+(1+x^2)dy=0$
Then:
$\frac{\partial \phi}{\partial x}=2xy-\frac{1}{1+x^2}$
Integrating both sides:
$\phi=(1+x^2)y-\arctan x+f(y)$
We have:
$\frac{\partial phi}{\partial y}=1+x^2+f'(y)=1+x^2$
Since $f'(y)=0 \rightarrow f(y)=C$
C is a constant integration
The solution is:
$C=(1+x^2)y-\arctan x$
or
$y= \frac{C+\arctan x}{1+x^2}$
