Expand the expression as
(s + 1)³/s ⁵ = (s ³ + 3s ² + 3s + 1)/s ⁵
… = 1/s ² + 3/s ³ + 3/s ⁴ + 1/s ⁵
Then taking the inverse transform, you get
LT⁻¹ [1/s ² + 3/s ³ + 3/s ⁴ + 1/s ⁵]
… = LT⁻¹ [1/s ²] + LT⁻¹ [3/s ³] + LT⁻¹ [3/s ⁴] + LT⁻¹ [1/s ⁵]
… = LT⁻¹ [1!/s ²] + 3/2 LT⁻¹ [2!/s ³] + 1/2 LT⁻¹ [3!/s ⁴] + 1/24 LT⁻¹ [4!/s ⁵]
… = t + 3/2 t ² + 1/2 t ³ + 1/24 t ⁴