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.

We multiply the ODE by the integrating factor $e^{\int{\frac{x}{x^2+1}}dx}=e^{\frac{ln(1+x^2)}{2}}=\sqrt{1+x^2}$ to obtain $\frac{d}{dx}(y\sqrt{1+x^2})=x\sqrt{1+x^2}$. Then integrating gives $y\sqrt{1+x^2}=\int x\sqrt{1+x^2}dx= \frac{(1+x^2)^{\frac{3}{2}}}{3}+C$ and finally we have $y(x)=\frac{C}{\sqrt{1+x^2}}+\frac{x^2+1}{3}$.
