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 :

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

 A: Yes this is solvable. Variation of parameters should do the trick.
A: If we use the method of Laplace transforms, denoting the Laplace transform of $f$ by $F$ and the Laplace transform of $1/(1+x^2)$ by $G$:
$$s^2 F(s) - s f(0) - f'(0) + F(s) = G(s)$$
and so
$$F(s) = \frac{G(s) + s f(0) + f'(0)}{s^2+1} \equiv \frac{H(s)}{s^2+1}$$
It is known (cf. Laplace transform tables) that the inverse Laplace transform of $\frac{1}{s^2+1}$ is $\sin(x)$. Then using the convolution property of the Laplace transform, we have $f(x)=h * \sin(x)$, where $h$ is the inverse Laplace transform of $H$ and $*$ denotes convolution. Linearity of the inverse Laplace transform gives $h(x) = \frac{1}{1+x^2} + \delta'(x) f(0) + \delta(x) f'(0)$, where $\delta$ is the Dirac delta. So we have the general solution
\begin{eqnarray*}
f(x) & = & \int_0^x \sin(x-y) \left ( \frac{1}{1+y^2} + \delta'(y) f(0) + \delta(y) f'(0) \right ) dy \\
& = & f(0) \cos(x) + f'(0) \sin(x) + \int_0^x \frac{\sin(x-y)}{1+y^2} dy 
\end{eqnarray*}
This last integral cannot be expressed in terms of elementary functions.
