# 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.

## Inverse Laplace of 1/s(s+1)

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.

## 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)}.