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

$\dfrac{1}{s(s+1)}$ = $\dfrac{s+1-s}{s(s+1)}$

= $\dfrac{s+1}{s(s+1)}$ – $\dfrac{s}{s(s+1)}$

= $\dfrac{1}{s}$ – $\dfrac{1}{s+1}$

Therefore,

$\dfrac{1}{s(s+1)}$ = $\dfrac{1}{s}$ – $\dfrac{1}{s+1}$

Taking inverse Laplace on both sides, we get that

L^{-1}$\big \{\dfrac{1}{s(s+1)} \big\}$ = L^{-1}$\big \{\dfrac{1}{s} – \dfrac{1}{s+1} \big\}$

= L^{-1} $\big \{\dfrac{1}{s} \big\}$ – L^{-1}$\big \{\dfrac{1}{s+1} \big\}$

= 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}.

Have You Read These?

- Table of Inverse Laplace Transformations
- Inverse Laplace transform of 1
- Inverse Laplace transform of 1/s
- Inverse Laplace transform of 1/s
^{2} - Inverse Laplace of 1/s
^{3}

## 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}.