1. **State the problem:** Find the inverse Laplace transform of the function $$F(s) = \frac{e^{-(s+1)}}{(s+1)(s^2 + 2s + 10)}.$$\n\n2. **Recall the shifting theorem:** The factor $e^{-as}$ in the Laplace domain corresponds to a time shift by $a$ in the time domain. Specifically, if $$\mathcal{L}\{f(t)\} = F(s),$$ then $$\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t - a)f(t - a),$$ where $u(t - a)$ is the Heaviside step function.\n\n3. **Rewrite the function:** Let $$G(s) = \frac{1}{(s+1)(s^2 + 2s + 10)}.$$ Then $$F(s) = e^{-(s+1)} G(s).$$ Note that $$e^{-(s+1)} = e^{-s} e^{-1}$$ but the shift applies to $s$ in the exponent, so the shift is by 1.\n\n4. **Focus on $G(s)$:** We want to find $$g(t) = \mathcal{L}^{-1}\{G(s)\} = \mathcal{L}^{-1}\left\{\frac{1}{(s+1)(s^2 + 2s + 10)}\right\}.$$\n\n5. **Complete the square in the quadratic:** $$s^2 + 2s + 10 = (s+1)^2 + 9.$$\n\n6. **Use partial fraction decomposition:** Set $$\frac{1}{(s+1)((s+1)^2 + 9)} = \frac{A}{s+1} + \frac{Bs + C}{(s+1)^2 + 9}.$$\n\n7. **Solve for coefficients:** Multiply both sides by the denominator: $$1 = A((s+1)^2 + 9) + (Bs + C)(s+1).$$ Substitute $s = -1$ to find $A$: $$1 = A(0 + 9) \Rightarrow A = \frac{1}{9}.$$\n\n8. **Expand and equate coefficients:**\n$$1 = \frac{1}{9}((s+1)^2 + 9) + (Bs + C)(s+1)$$\n$$= \frac{1}{9}(s^2 + 2s + 10) + (Bs + C)(s+1).$$\n\nMultiply out $(Bs + C)(s+1) = Bs^2 + Bs + Cs + C = Bs^2 + (B + C)s + C.$\n\nSo,\n$$1 = \frac{1}{9}s^2 + \frac{2}{9}s + \frac{10}{9} + Bs^2 + (B + C)s + C.$$\n\nGroup terms:\n$$1 = (B + \frac{1}{9}) s^2 + (B + C + \frac{2}{9}) s + (C + \frac{10}{9}).$$\n\nSince the left side is constant 1, coefficients of $s^2$ and $s$ must be zero:\n$$B + \frac{1}{9} = 0 \Rightarrow B = -\frac{1}{9},$$\n$$B + C + \frac{2}{9} = 0 \Rightarrow -\frac{1}{9} + C + \frac{2}{9} = 0 \Rightarrow C = -\frac{1}{9}.$$\n\nConstant term:\n$$C + \frac{10}{9} = 1 \Rightarrow -\frac{1}{9} + \frac{10}{9} = 1,$$ which is true.\n\n9. **Rewrite partial fractions:**\n$$G(s) = \frac{1}{9} \cdot \frac{1}{s+1} - \frac{1}{9} \cdot \frac{s}{(s+1)^2 + 9} - \frac{1}{9} \cdot \frac{1}{(s+1)^2 + 9}.$$\n\n10. **Recall inverse Laplace transforms:**\n- $$\mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\} = e^{-at}.$$\n- $$\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2 + b^2}\right\} = e^{-at} \cos(bt).$$\n- $$\mathcal{L}^{-1}\left\{\frac{b}{(s+a)^2 + b^2}\right\} = e^{-at} \sin(bt).$$\n\n11. **Rewrite terms to match formulas:**\n$$\frac{s}{(s+1)^2 + 9} = \frac{(s+1) - 1}{(s+1)^2 + 9} = \frac{s+1}{(s+1)^2 + 9} - \frac{1}{(s+1)^2 + 9}.$$\n\nSo,\n$$G(s) = \frac{1}{9} \cdot \frac{1}{s+1} - \frac{1}{9} \left( \frac{s+1}{(s+1)^2 + 9} - \frac{1}{(s+1)^2 + 9} \right) - \frac{1}{9} \cdot \frac{1}{(s+1)^2 + 9}.$$\n\nSimplify:\n$$G(s) = \frac{1}{9} \cdot \frac{1}{s+1} - \frac{1}{9} \cdot \frac{s+1}{(s+1)^2 + 9} + \frac{1}{9} \cdot \frac{1}{(s+1)^2 + 9} - \frac{1}{9} \cdot \frac{1}{(s+1)^2 + 9} = \frac{1}{9} \cdot \frac{1}{s+1} - \frac{1}{9} \cdot \frac{s+1}{(s+1)^2 + 9}.$$\n\n12. **Inverse Laplace transform of $G(s)$:**\n$$g(t) = \frac{1}{9} e^{-t} - \frac{1}{9} e^{-t} \cos(3t) = \frac{e^{-t}}{9} (1 - \cos(3t)).$$\n\n13. **Apply the time shift:** Since $$F(s) = e^{-(s+1)} G(s) = e^{-s} e^{-1} G(s),$$ the shift is by 1, so\n$$f(t) = u(t-1) g(t-1) = u(t-1) \frac{e^{-(t-1)}}{9} (1 - \cos(3(t-1))).$$\n\n**Final answer:**\n$$\boxed{f(t) = \frac{u(t-1)}{9} e^{-(t-1)} \left(1 - \cos(3(t-1))\right)}.$$