1. The problem is to find the integral of the function $$\frac{\cos x}{D^2+1}$$ where $D$ is the differential operator $\frac{d}{dx}$. This is interpreted as solving the differential equation or finding the integral involving the operator $D^2+1$.\n\n2. The operator $D^2+1$ corresponds to the differential equation $y'' + y = f(x)$ where $f(x) = \cos x$.\n\n3. To solve $y'' + y = \cos x$, we find the complementary solution $y_c$ and a particular solution $y_p$.\n\n4. The complementary solution solves $y'' + y = 0$, whose characteristic equation is $r^2 + 1 = 0$ with roots $r = \pm i$. Thus, \n$$y_c = C_1 \cos x + C_2 \sin x.$$\n\n5. For the particular solution $y_p$, since the right side is $\cos x$, which is a solution to the homogeneous equation, we try a particular solution of the form \n$$y_p = x(A \sin x + B \cos x).$$\n\n6. Compute derivatives:\n$$y_p' = A \sin x + B \cos x + x(A \cos x - B \sin x),$$\n$$y_p'' = 2A \cos x - 2B \sin x - x(A \sin x + B \cos x).$$\n\n7. Substitute into the differential equation:\n$$y_p'' + y_p = (2A \cos x - 2B \sin x - x(A \sin x + B \cos x)) + x(A \sin x + B \cos x) = 2A \cos x - 2B \sin x.$$\n\n8. Set equal to $\cos x$:\n$$2A \cos x - 2B \sin x = \cos x.$$\n\n9. Equate coefficients:\n$$2A = 1 \Rightarrow A = \frac{1}{2}, \quad -2B = 0 \Rightarrow B = 0.$$\n\n10. Thus, the particular solution is \n$$y_p = \frac{1}{2} x \sin x.$$\n\n11. The general solution is \n$$y = y_c + y_p = C_1 \cos x + C_2 \sin x + \frac{1}{2} x \sin x.$$\n\n12. Therefore, the integral or solution corresponding to $\frac{\cos x}{D^2+1}$ is \n$$\boxed{\frac{1}{2} x \sin x}.$$