A second-order equation with a polynomial right-hand side

Homogeneous equation. $r^2 + 9 = 0$ gives $r = \pm 3i$, so
$$y_h(x) = A\cos(3x) + B\sin(3x).$$

Particular solution. Try $y_p = ax+b$. Then $y_p'' = 0$, and
$$y_p'' + 9y_p = 9(ax+b) = 9ax + 9b.$$
Identifying with $x+1$:
$$9a = 1 \implies a = \frac19, \qquad 9b = 1 \implies b = \frac19.$$
So
$$y_p(x) = \frac{x+1}{9}.$$

General solution.
$$y(x) = A\cos(3x) + B\sin(3x) + \frac{x+1}{9}.$$