A Table Of Laplace Transforms going from f(s) to F(t) shows (1/s3) transformed to t2/2! and 1/(s-1) transformed to e(1)t or et.
Then obtain, by Convolution, L-1{1/(s3(s-1)} equal to ∫(from u=0 to u=t)(u2/2!)e(t-u)du.
Rewrite ∫(from u=0 to u=t)(u2/2!)e(t-u)du as (et/2!)∫(from u=0 to u=t)u2e-udu.
Integration By Parts will give (et/2!)∫(from u=0 to u=t)u2e-udu as (et/2!)[-u2e-u −2ue-u − 2e-u|(from u=0 to u=t)] which equals (et/2)[-t2e-t − 2te-t − 2e-t + 2] or -t2/2 − t − 1 + et.