Evaluate the inverse laplace transform of the function f(s)=1/s+42-4s/(s+3)^4
f(s) = 1/(s + 42) - 4s/(s + 3)^4
L^(-1){1/(s + 42)} = exp(-42t)
s/(s + 3)^4 = {(s + 3) - 3} / (s + 3)^4
s/(s + 3)^4 = (s + 3)/(s + 3)^4 - 3/(s + 3)^4
s/(s + 3)^4 = 1/(s + 3)^3 - 3/(s + 3)^4
s/(s + 3)^4 = (1/2)*2!/(s + 3)^(2+1) - (1/2)*3!/(s + 3)^(3+1)
L^(-1){s/(s + 3)^4} = (1/2)*t^2.exp(-3t) - (1/2)*t^3.exp(-3t)
L^(-1){s/(s + 3)^4} = (1/2)*t^2.exp(-3t){1 - t)
Finally,
L^(-1){1/(s + 42) - 4s/(s + 3)^4} = L^(-1){1/(s + 42)} - 4L^(-1){s/(s + 3)^4}
L^(-1){1/(s + 42) - 4s/(s + 3)^4} = exp(-42t) - 2.t^2.exp(-3t){1 - t)
f(t) = exp(-42t) - 2.t^2.exp(-3t){1 - t)