Question

(2s+5)/(s^2+6s+34) inverse Laplace transform

Answer 1

(2s+5)/(s^2+6s+9+25) = (2s+6-1)/((s+3)^2+25) = (2(s+3)-1)/((s+3)^2+5^2) = 2[(s+3)/((s+3)^2+5^2) - 1/5 (5 / ((s+3)^2+5^2))
Hence inverse Laplace transform is 2e^(-3t)cos 5t - (1/5) e^(-3t)sin 5t