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.

This is a linear ODE then
$$
y=y_h+y_p\\
y'_h + \frac{x}{1+x^2}y_h = 0\\
y'_p + \frac{x}{1+x^2}y_p = x
$$
the homogeneous is separable giving
$$
y_h = \frac{c_0}{\sqrt{1+x^2}}
$$
now using the method of constants variation due to Lagrange we make $y_p = \frac{c_0(x)}{\sqrt{1+x^2}}$ and substituting we obtain
$$
\frac{c_0'(x)}{\sqrt{x^2+1}}-x=0
$$
giving
$$
c_0(x) = \frac{1}{3} \left(x^2+1\right)^{3/2}
$$
and finally
$$
y = \frac{c_0}{\sqrt{1+x^2}}+\frac{1}{3} \left(x^2+1\right)^{3/2}\frac{1}{\sqrt{1+x^2}} = \frac{c_0}{\sqrt{1+x^2}}+\frac{1}{3} \left(x^2+1\right)
$$
