Right option is (b) \(\frac{(s+3)}{s[(s+3)^2+4]}\)
For explanation: Laplace transform of e-at cost u (t) = \(\frac{s+a}{(s+a)^2+1}\)
Let p (t) = e^-3t cos2t u (t)
Laplace transform of p (t), p(s) = \(\frac{s+3}{(s+3)^2+4}\)
∴ Laplace transform of \(\int_{-∞}^t \,p(t)dt = \frac{1}{s} \int_{-∞}^0 \,p(t)dt + \frac{p(s)}{s}\)
Or, X(s) = \(\frac{(s+3)}{s[(s+3)^2+4]}\).