Correct answer is (b) \(\frac{s^2+2s+16}{s(s^2+4^2)}\)
Explanation: Given x(t)=[1 + sin 2t cos 2t]u(t) = (1 + \(\frac{1}{2} \,sin4t\))u(t)
L{x(t)} = X(s) = L[u(t) + \(\frac{1}{2}\) sin4t u(t)] = L[u(t)] + \(\frac{1}{2}\) L[sin4t u(t)] = \(\frac{1}{s} + \frac{1}{2} \frac{4}{(s^2+4^2)} = \frac{s^2+2s+16}{s(s^2+4^2)}\).