1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram 
$$y=\frac{1}{\sqrt{x^2+1}}\cdot C$$
But when I try to calculate $y'$, I get a strange equation. I think I had to be wrong somewhere. I will be grateful for your help.

$$y'+\frac{xy}{1+x^2} =x$$
$$2x\dfrac {dy}{dx^2}+\frac{xy}{1+x^2} =x$$
Substitute $u=x^2$
$$2\dfrac {dy}{du}+\frac{y}{1+u} =1$$
$$2\sqrt {1+u}{y'}+\frac{y}{\sqrt {1+u}}=\sqrt {1+u}$$
$$(2\sqrt {1+u}{y})'=\sqrt {1+u}$$
Integrate:
$$2\sqrt {1+u}{y}=\int \sqrt {1+u}du$$
$$\sqrt {1+u}{y}=\frac 13({1+u})^{3/2}+K$$
$${y}=\frac 13({1+x^2})+\dfrac K {\sqrt {1+x^2}}$$
