Inverse Laplace Transform of 1/(s + 2)

The inverse Laplace transform of 1/(s + 2) is e^(-2t). 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 + 2)} = e^(-2t). A pole at −2 gives the decaying exponential e^(−2t).

Numerator N(s)

e.g. 2s+3

Denominator D(s)

Expanded or factored, e.g. (s+1)(s^2+4)

Try an example

The result

L⁻¹{1/(s + 2)} = e^(-2t)

Partial fractions: 1/(s + 2)

Invert 1/(s + 2) — L⁻¹{A/(s − a)^1} = A·t^0e^(at)/0!

A pole at −2 gives the decaying exponential e^(−2t).

More inverse transforms

L⁻¹{1/s} L⁻¹{1/s²} L⁻¹{1/s³} L⁻¹{1/(s − 3)} L⁻¹{1/(s² + 4)} L⁻¹{s/(s² + 9)} L⁻¹{1/(s² − 4)} L⁻¹{1/(s(s + 1))}

Frequently asked questions

What is the inverse Laplace transform of 1/(s + 2)? ▾

It is f(t) = e^(-2t). A pole at −2 gives the decaying exponential e^(−2t).

What are the partial fractions here? ▾

F(s) splits into 1/(s + 2), and each piece is inverted separately.

How can I check this answer? ▾

Transform e^(-2t) forward again with the Laplace transform calculator — you should get back 1/(s + 2).

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