Inverse Laplace Transform of 1/(s² + 1)²
The inverse Laplace transform of 1/(s² + 1)² is 1/2(sin(t) − t·cos(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/(s² + 1)²} = 1/2(sin(t) − t·cos(t)). A repeated quadratic factor produces the resonance term sin t − t cos t.
e.g. 2s+3
Expanded or factored, e.g. (s+1)(s^2+4)
The result
L⁻¹{1/(s² + 1)²} = 1/2(sin(t) − t·cos(t))
Partial fractions: 1/(s^2 + 1)^2
- Invert 1/(s^2 + 1)^2 — L⁻¹{1/(s² + b²)²} = (sin bt − bt·cos bt)/(2b³)
A repeated quadratic factor produces the resonance term sin t − t cos t.
More inverse transforms
Frequently asked questions
What is the inverse Laplace transform of 1/(s² + 1)²? ▾
It is f(t) = 1/2(sin(t) − t·cos(t)). A repeated quadratic factor produces the resonance term sin t − t cos t.
What are the partial fractions here? ▾
F(s) splits into 1/(s^2 + 1)^2, and each piece is inverted separately.
How can I check this answer? ▾
Transform 1/2(sin(t) − t·cos(t)) forward again with the Laplace transform calculator — you should get back 1/(s² + 1)².
About this Calculator
The inverse Laplace transform of 1/(s² + 1)² is 1/2(sin(t) − t·cos(t)). See the partial fraction expansion and the transform pair used for each term, then edit F(s) to try your own.