1. We are asked to find the integral $$\int x \sin(mx) \, dx$$ where $$m \neq 0$$. 2. The formula for integration by parts is $$\int u \, dv = uv - \int v \, du$$. 3. Let $$u = x$$ and $$dv = \sin(mx) \, dx$$. 4. Then, $$du = dx$$ and $$v = -\frac{\cos(mx)}{m}$$ (since $$\int \sin(mx) \, dx = -\frac{\cos(mx)}{m}$$). 5. Applying integration by parts: $$\int x \sin(mx) \, dx = u v - \int v \, du = -\frac{x \cos(mx)}{m} - \int -\frac{\cos(mx)}{m} \, dx$$ 6. Simplify the integral: $$= -\frac{x \cos(mx)}{m} + \frac{1}{m} \int \cos(mx) \, dx$$ 7. Calculate $$\int \cos(mx) \, dx = \frac{\sin(mx)}{m}$$. 8. Substitute back: $$= -\frac{x \cos(mx)}{m} + \frac{1}{m} \cdot \frac{\sin(mx)}{m} + C = -\frac{x \cos(mx)}{m} + \frac{\sin(mx)}{m^2} + C$$ 9. Therefore, the integral is: $$\boxed{-\frac{x \cos(mx)}{m} + \frac{\sin(mx)}{m^2} + C}$$ This matches the first option.