Let's solve the given inverse Laplace transform step-by-step: Given: \[ \mathcal{L}^{-1}\left\{ \frac{8s-16}{(s^2 + s)(s^2 + 1)} \right\} \] ### **Step 1: Factor and Set Up Partial Fractions** First, factor the denominator: \[ (s^2 + s)(s^2 + 1) = s(s + 1)(s^2 + 1) \] Write the expression as partial fractions: \[ \frac{8s-16}{s(s+1)(s^2+1)} = \frac{A}{s} + \frac{B}{s+1} + \frac{Cs+D}{s^2+1} \] ### **Step 2: Find the Constants** Multiply both sides by the denominator: \[ 8s-16 = A(s+1)(s^2+1) + B s(s^2+1) + (Cs+D)s(s+1) \] Expand: - \(A(s+1)(s^2+1) = A(s+1)(s^2+1) = A(s^3 + s^2 + s + 1)\) - \(B s(s^2+1) = B(s^3 + s)\) - \((Cs+D)s(s+1) = Cs^2(s+1) + Ds(s+1) = Cs^3 + Cs^2 + Ds^2 + Ds\) So, \[ 8s-16 = A(s^3 + s^2 + s + 1) + B(s^3 + s) + C(s^3 + s^2) + D(s^2 + s) \] Group like powers: \[ 8s-16 = (A+B+C)s^3 + (A+C+D)s^2 + (A+B+D)s + A \] ### **Step 3: Set Up Equations for Coefficients** Match coefficients with \(8s-16\): - \(s^3\): \(A + B + C = 0\) - \(s^2\): \(A + C + D = 0\) - \(s^1\): \(A + B + D = 8\) - constant: \(A = -16\) From here: 1. \(A = -16\) 2. \(A + B + C = 0 \implies -16 + B + C = 0 \implies B + C = 16\) 3. \(A + C + D = 0 \implies -16 + C + D = 0 \implies C + D = 16\) 4. \(A + B + D = 8 \implies -16 + B + D = 8 \implies B + D = 24\) Now solve this system: From (2): \(B + C = 16\) From (3): \(C + D = 16\) From (4): \(B + D = 24\) Let’s solve for \(B, C, D\): From (2): \(B = 16 - C\) From (3): \(D = 16 - C\) Plug into (4): \(B + D = 24\): \[ (16 - C) + (16 - C) = 24 \implies 32 - 2C = 24 \implies C = 4 \] Now, \(B = 16 - 4 = 12\) \(D = 16 - 4 = 12\) So, \[ A = -16, \quad B = 12, \quad C = 4, \quad D = 12 \] ### **Step 4: Write Out the Partial Fractions** \[ \frac{8s-16}{s(s+1)(s^2+1)} = \frac{-16}{s} + \frac{12}{s+1} + \frac{4s+12}{s^2+1} \] ### **Step 5: Take the Inverse Laplace Transform** Recall: - \(\mathcal{L}^{-1}\left\{ \frac{1}{s} \right\} = 1\) - \(\mathcal{L}^{-1}\left\{ \frac{1}{s+a} \right\} = e^{-at}\) - \(\mathcal{L}^{-1}\left\{ \frac{s}{s^2+1} \right\} = \cos t\) - \(\mathcal{L}^{-1}\left\{ \frac{1}{s^2+1} \right\} = \sin t\) So, \[ \mathcal{L}^{-1}\left\{ \frac{-16}{s} \right\} = -16 \] \[ \mathcal{L}^{-1}\left\{ \frac{12}{s+1} \right\} = 12 e^{-t} \] \[ \mathcal{L}^{-1}\left\{ \frac{4s}{s^2+1} \right\} = 4\cos t \] \[ \mathcal{L}^{-1}\left\{ \frac{12}{s^2+1} \right\} = 12\sin t \] ### **Final Answer** \[ \boxed{ f(t) = -16 + 12e^{-t} + 4\cos t + 12\sin t } \]