Solving $(1+x^2)y' - 2xy = (1+x^2)\arctan(x)$ I'm asked to solve the differential equation: $$(1+x^2)y' - 2xy = (1+x^2)\arctan(x).$$
I rewrite it: 
$$y' - \frac{2x}{1+x^2}y = \arctan(x).$$
The integrating factor is: 
$$e^{-\int{\frac{2x}{1+x^2}}dx} = e^{-\ln{1+x^2}} = (e^{ln{1+x^2}})^{-1} = \frac{1}{1+x^2}.$$ 
$$\frac{1}{1+x^2}y' - \frac{1}{1+x^2}\frac{2x}{1+x^2}y = \arctan(x)\frac{1}{1+x^2}$$
$$\frac{d}{dx}(\frac{1}{1+x^2}y) = \arctan(x)\frac{1}{1+x^2}$$
So to solve this equation I need to find the primitive of 
$$\arctan(x)\frac{1}{1+x^2}$$
But I don't think I can do that yet, is there any other way to solve this differential equation? 

The derivative of $\arctan$ is $x\mapsto \frac{1}{1+x^{2}}$.
Now, observe that for a differentiable function $f$, we have:
$$\int f(x)f'(x)\text{d}x=\frac{1}{2}(f(x))^{2}+K$$
where $K\in\mathbb{R}$. Taking $f=\arctan$, we deduce what you asked.
