Integrate the function: $x \sin 3x$.

(N/A) Let $I = \int x \sin 3x \, dx$.
Using the integration by parts formula: $\int u v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$.
Taking $u = x$ (first function) and $v = \sin 3x$ (second function),we get:
$I = x \int \sin 3x \, dx - \int \left( \frac{d}{dx}(x) \int \sin 3x \, dx \right) dx$
$I = x \left( \frac{-\cos 3x}{3} \right) - \int 1 \cdot \left( \frac{-\cos 3x}{3} \right) dx$
$I = -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx$
$I = -\frac{x \cos 3x}{3} + \frac{1}{3} \left( \frac{\sin 3x}{3} \right) + C$
$I = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C$,where $C$ is an arbitrary constant.