Inverse Laplace Transform of 1/(s(s + 1))
The inverse Laplace transform of 1/(s(s + 1)) is 1 - e^(-t). See the partial fraction expansion and the transform pair used for each term, then edit F(s) to try your own.
L⁻¹{1/(s(s + 1))} = 1 - e^(-t). The step response of a first-order system: 1 − e^(−t).
e.g. 2s+3
Expanded or factored, e.g. (s+1)(s^2+4)
The result
L⁻¹{1/(s(s + 1))} = 1 - e^(-t)
Partial fractions: 1/s + -1/(s + 1)
- Invert 1/s — L⁻¹{A/(s − a)^1} = A·t^0e^(at)/0!
- Invert -1/(s + 1) — L⁻¹{A/(s − a)^1} = A·t^0e^(at)/0!
The step response of a first-order system: 1 − e^(−t).
More inverse transforms
Frequently asked questions
What is the inverse Laplace transform of 1/(s(s + 1))? ▾
It is f(t) = 1 - e^(-t). The step response of a first-order system: 1 − e^(−t).
What are the partial fractions here? ▾
F(s) splits into 1/s + -1/(s + 1), and each piece is inverted separately.
How can I check this answer? ▾
Transform 1 - e^(-t) forward again with the Laplace transform calculator — you should get back 1/(s(s + 1)).
About this Calculator
The inverse Laplace transform of 1/(s(s + 1)) is 1 - e^(-t). See the partial fraction expansion and the transform pair used for each term, then edit F(s) to try your own.