1. **State the problem:** Find the inverse Laplace transform of $$\frac{3s+2}{s^2+4s+16}$$. 2. **Rewrite the denominator:** Complete the square for the quadratic in the denominator: $$s^2 + 4s + 16 = (s^2 + 4s + 4) + 12 = (s+2)^2 + 12$$ 3. **Rewrite the numerator to match the form:** We want to express $$3s + 2$$ in terms of $$s+2$$: $$3s + 2 = 3(s+2) - 4$$ 4. **Rewrite the fraction:** $$\frac{3s+2}{s^2+4s+16} = \frac{3(s+2) - 4}{(s+2)^2 + 12} = 3 \cdot \frac{s+2}{(s+2)^2 + (2\sqrt{3})^2} - 4 \cdot \frac{1}{(s+2)^2 + (2\sqrt{3})^2}$$ 5. **Recall Laplace inverse formulas:** - $$\mathcal{L}^{-1}\left[\frac{s-a}{(s-a)^2 + b^2}\right] = e^{at} \cos(bt)$$ - $$\mathcal{L}^{-1}\left[\frac{b}{(s-a)^2 + b^2}\right] = e^{at} \sin(bt)$$ 6. **Adjust the second term to match the sine form:** We have $$\frac{1}{(s+2)^2 + (2\sqrt{3})^2}$$ but need $$\frac{2\sqrt{3}}{(s+2)^2 + (2\sqrt{3})^2}$$ for the sine formula. Multiply numerator and denominator by $$2\sqrt{3}$$ and compensate: $$\frac{1}{(s+2)^2 + (2\sqrt{3})^2} = \frac{2\sqrt{3}}{(s+2)^2 + (2\sqrt{3})^2} \cdot \frac{1}{2\sqrt{3}}$$ 7. **Rewrite the expression:** $$3 \cdot \frac{s+2}{(s+2)^2 + (2\sqrt{3})^2} - 4 \cdot \frac{1}{(s+2)^2 + (2\sqrt{3})^2} = 3 \cdot \frac{s+2}{(s+2)^2 + (2\sqrt{3})^2} - \frac{4}{2\sqrt{3}} \cdot \frac{2\sqrt{3}}{(s+2)^2 + (2\sqrt{3})^2}$$ 8. **Apply inverse Laplace transform:** $$\mathcal{L}^{-1}\left[\frac{3s+2}{s^2+4s+16}\right] = 3 e^{-2t} \cos(2\sqrt{3} t) - \frac{4}{2\sqrt{3}} e^{-2t} \sin(2\sqrt{3} t)$$ 9. **Simplify the coefficient:** $$\frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$$ 10. **Final answer:** $$\boxed{f(t) = e^{-2t} \left(3 \cos(2\sqrt{3} t) - \frac{2}{\sqrt{3}} \sin(2\sqrt{3} t)\right)}$$