Given F(s) = 1/((s-1)^(2)(s+1))

Question

After using partial fractions, I got (1/4)[(2/((s-1)^(2)))+(1/(s+1))-(1/(s-1))] before applying Laplace inverse.

Apparently, the answer is (1/4)[2te^(t) + e^(-t) - e^(t)], however I don't know where the t in 2te^(t) came from. Can someone please explain? Thanks in advance! :D

Answer 1

Recall that L{tf(t)) = -F'(s)

Since L^-1{1/(s-1)} = e^t
and d/ds 1/(s-1) = 1/(s-1)^2

L^-1{1/(s-1)^2} = te^t