1. **State the problem:** We need to evaluate the definite integral $$\int_{-\sqrt{6}}^{\pi} (2x + 2|x|) \sin(6x) \, dx.$$\n\n2. **Understand the absolute value:** The function inside the integral involves $|x|$, which behaves differently for $x<0$ and $x\geq0$.\n- For $x<0$, $|x| = -x$.\n- For $x\geq0$, $|x| = x$.\n\n3. **Split the integral at 0:** Because of the absolute value, split the integral into two parts:\n$$\int_{-\sqrt{6}}^{0} (2x + 2|x|) \sin(6x) \, dx + \int_{0}^{\pi} (2x + 2|x|) \sin(6x) \, dx.$$\n\n4. **Simplify each part:**\n- For $x<0$, $|x| = -x$, so inside the integral: $$2x + 2|x| = 2x + 2(-x) = 2x - 2x = 0.$$\n- For $x\geq0$, $|x| = x$, so inside the integral: $$2x + 2|x| = 2x + 2x = 4x.$$\n\n5. **Evaluate the first integral:** Since the integrand is 0 for $x<0$,\n$$\int_{-\sqrt{6}}^{0} 0 \cdot \sin(6x) \, dx = 0.$$\n\n6. **Evaluate the second integral:**\n$$\int_{0}^{\pi} 4x \sin(6x) \, dx = 4 \int_{0}^{\pi} x \sin(6x) \, dx.$$\n\n7. **Use integration by parts:** Let $u = x$, $dv = \sin(6x) dx$. Then $du = dx$, and $v = -\frac{\cos(6x)}{6}$.\n\nIntegration by parts formula: $$\int u \, dv = uv - \int v \, du.$$\n\nSo,\n$$\int x \sin(6x) dx = -\frac{x \cos(6x)}{6} + \frac{1}{6} \int \cos(6x) dx.$$\n\n8. **Integrate $\int \cos(6x) dx$:**\n$$\int \cos(6x) dx = \frac{\sin(6x)}{6} + C.$$\n\n9. **Combine results:**\n$$\int x \sin(6x) dx = -\frac{x \cos(6x)}{6} + \frac{\sin(6x)}{36} + C.$$\n\n10. **Evaluate definite integral from 0 to $\pi$:**\n$$\int_0^{\pi} x \sin(6x) dx = \left[-\frac{x \cos(6x)}{6} + \frac{\sin(6x)}{36}\right]_0^{\pi}.$$\n\nCalculate at $x=\pi$:\n$$-\frac{\pi \cos(6\pi)}{6} + \frac{\sin(6\pi)}{36} = -\frac{\pi \cdot 1}{6} + 0 = -\frac{\pi}{6}.$$\n\nCalculate at $x=0$:\n$$-\frac{0 \cdot \cos(0)}{6} + \frac{\sin(0)}{36} = 0 + 0 = 0.$$\n\nSo,\n$$\int_0^{\pi} x \sin(6x) dx = -\frac{\pi}{6} - 0 = -\frac{\pi}{6}.$$\n\n11. **Multiply by 4:**\n$$4 \times \left(-\frac{\pi}{6}\right) = -\frac{2\pi}{3}.$$\n\n12. **Final answer:**\n$$\int_{-\sqrt{6}}^{\pi} (2x + 2|x|) \sin(6x) \, dx = 0 + \left(-\frac{2\pi}{3}\right) = -\frac{2\pi}{3}.$$