Ordinary Differential Equation with trigs! I need help and alternative ways of solving this problem: 
$$ y'(x^2 + 1) - 2xy = (x^2 + 1)\arctan x $$
Thank you for all kind of help and hints :) 

$$
\left(x^2+1\right)y' -2xy = \left(x^2+1\right)\arctan x 
$$
$$
y' -\frac{2x}{x^2+1}y = \arctan x 
$$
$$
\frac{g'}{g} = \frac{2x}{x^2+1} = \frac{d}{dx}\ln g(x)
$$
thus
$$
y\mathrm{e}^{-\ln g} = \frac{y}{g} = \int \frac{1}{g}\arctan x dx = \int \frac{1}{x^2+1}\arctan x dx 
$$
$$
\frac{d}{dx}\arctan x = \frac{1}{x^2+1}
$$
therefore
$$
\frac{y}{x^2+1} = \int \left(\frac{d}{dx}\arctan x\right)\arctan x dx = \int \frac{d}{dx}\frac{\arctan^2 x}{2} dx = \frac{1}{2}\arctan^2(x) +C.
$$
