second degree differential equation Find all functions $f(x)$ such that 

$$f''(x)+f(x)=\frac{1}{1+x^2}.$$

I would like to know if it's solvable and the solution/hints.
What I got :

$$2f(x)f(x)'+2f(x)'f(x)''=2\frac{1}{1+x^2}f'(x)$$
  $$(f(x)^2)'+((f(x)')^2)'=2\frac{1}{1+x^2}f'(x)$$
  $$f(x)^2+(f(x)')^2=\int 2\frac{1}{1+x^2}f'(x)dx .$$


Yes this is solvable. Variation of parameters should do the trick.
