1. **State the problem:** Calculate the inverse Laplace transform $x(t)$ given by $$x(t) = \mathcal{L}^{-1} \left\{ \frac{5}{2(s+1)} \right\} - \mathcal{L}^{-1} \left\{ \frac{6}{5(s+2)} \right\} + \mathcal{L}^{-1} \left\{ \frac{-3s+1}{10(s^2+1)} \right\}$$ 2. **Recall the inverse Laplace transform formulas:** - $\mathcal{L}^{-1} \left\{ \frac{1}{s+a} \right\} = e^{-at}$ - $\mathcal{L}^{-1} \left\{ \frac{s}{s^2+\omega^2} \right\} = \cos(\omega t)$ - $\mathcal{L}^{-1} \left\{ \frac{\omega}{s^2+\omega^2} \right\} = \sin(\omega t)$ 3. **Apply linearity and constants:** - $\mathcal{L}^{-1} \left\{ \frac{5}{2(s+1)} \right\} = \frac{5}{2} e^{-t}$ - $\mathcal{L}^{-1} \left\{ \frac{6}{5(s+2)} \right\} = \frac{6}{5} e^{-2t}$ 4. **Decompose the third term:** $$\frac{-3s+1}{10(s^2+1)} = -\frac{3}{10} \cdot \frac{s}{s^2+1} + \frac{1}{10} \cdot \frac{1}{s^2+1}$$ 5. **Use inverse transforms for cosine and sine:** - $\mathcal{L}^{-1} \left\{ \frac{s}{s^2+1} \right\} = \cos t$ - $\mathcal{L}^{-1} \left\{ \frac{1}{s^2+1} \right\} = \sin t$ 6. **Combine all results:** $$x(t) = \frac{5}{2} e^{-t} - \frac{6}{5} e^{-2t} - \frac{3}{10} \cos t + \frac{1}{10} \sin t$$ This is the solution for $x(t)$ using the given $X(s)$.