Inverse Laplace transform of 1/s(s+1)

The inverse Laplace transform of 1/s(s+1) is 1 -e-t, and it is denoted by L-1{1/s(s+1)}. Here we learn how to find the inverse Laplace of 1 divided by s(s+1).

The formula of the inverse Laplace of 1/s(s+1) is given below:

L-1{1/s(s+1)} = 1-e-t.

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Inverse Laplace of 1/s(s+1)

Answer: L-1{1/s(s+1)} = 1 -e-t.

Explanation:

Note that

1s(s+1) = s+1ss(s+1)

= s+1s(s+1)ss(s+1)

= 1s1s+1

Therefore,

1s(s+1) = 1s1s+1

Taking inverse Laplace on both sides, we get that

L-1{1s(s+1)} = L-1{1s1s+1}

= L-1 {1s} – L-1{1s+1}

= 1 – e-t, here we have used the inverse Laplace formula L-1{1/(s+a)} = e-at.

So the inverse Laplace of 1/s(s+1) is equal to 1 -e-t.

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FAQs

Q1: What is the inverse Laplace transform of 1/s(s+1)?

Answer: The inverse Laplace transform of 1/s(s+1) is equal to 1 -e-t, that is, L-1{1/s(s+1)} = 1-e-t.

Q2: Find L-1{1/s(s+1)}.

Answer: L-1{1/s(s+1)} = 1-e-t.

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