$\int_0^{\frac{\pi}{6}}\left(2+3 x^2\right) \cos 3 x d x=$

Evaluate the definite integral: $\int_0^{\frac{\pi}{6}} (2+3x^2)\cos 3x\,dx$

Apply integration by parts with $u = 2+3x^2$ and $dv = \cos 3x\,dx$, yielding $du = 6x\,dx$ and $v = \frac{1}{3}\sin 3x$:

$I = \frac{1}{3}(2+3x^2)\sin 3x - 2\int x\sin 3x\,dx$

Apply integration by parts again to the remaining integral with $u = x$ and $dv = \sin 3x\,dx$, giving $du = dx$ and $v = -\frac{1}{3}\cos 3x$:

$\int x\sin 3x\,dx = -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x$

Substitute back into the expression for $I$:

$I = \frac{1}{3}(2+3x^2)\sin 3x + \frac{2}{3}x\cos 3x - \frac{2}{9}\sin 3x$

Evaluate the definite integral from $0$ to $\frac{\pi}{6}$:

At $x = \frac{\pi}{6}$:
$\sin\frac{\pi}{2} = 1$, $\cos\frac{\pi}{2} = 0$
$\frac{1}{3}(2+3(\frac{\pi}{6})^2)(1) + \frac{2}{3}(\frac{\pi}{6})(0) - \frac{2}{9}(1) = \frac{4}{9} + \frac{\pi^2}{36}$

At $x = 0$, the expression evaluates to $0$.

Final result: $\frac{4}{9} + \frac{\pi^2}{36}$