Right option is (c) \(\frac{1}{s+4} – \frac{2}{s+2} + \frac{1}{s}\)
Easy explanation: (e^-2t – 1)^2 = e^-4t – 2e^-2t + 1
We know that, L {e^-at} = \(\frac{1}{s+a}\)
L {1} = \(\frac{1}{s}\)
∴L {(e^-2t – 1)^2} = L {e^-4t – 2e^-2t + 1} = \(\frac{1}{s+4} – \frac{2}{s+2} + \frac{1}{s}\).