1. **State the problem:** Find the inverse Laplace transform of the expression $$\sqrt{5-1} + \ln \sqrt{8^2 - 1}$$. 2. **Simplify the expression inside the Laplace inverse:** Calculate the square root and logarithm terms: $$\sqrt{5-1} = \sqrt{4} = 2$$ $$\sqrt{8^2 - 1} = \sqrt{64 - 1} = \sqrt{63}$$ So the expression becomes: $$2 + \ln \sqrt{63}$$ 3. **Simplify the logarithm:** Recall that $$\ln \sqrt{63} = \ln 63^{1/2} = \frac{1}{2} \ln 63$$ Thus, the expression is: $$2 + \frac{1}{2} \ln 63$$ 4. **Interpretation:** The expression inside the Laplace inverse is a constant (no variable $s$). The Laplace transform of a constant $c$ is $$\frac{c}{s}$$. Therefore, the inverse Laplace transform of a constant $c$ is: $$\mathcal{L}^{-1}[c] = c \delta(t)$$ where $\delta(t)$ is the Dirac delta function. 5. **Final answer:** $$\mathcal{L}^{-1} \left[ 2 + \frac{1}{2} \ln 63 \right] = \left( 2 + \frac{1}{2} \ln 63 \right) \delta(t)$$ This means the inverse Laplace transform is a scaled delta function at $t=0$ with scale factor $$2 + \frac{1}{2} \ln 63$$.