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