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)

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/s²
• Inverse Laplace of 1/s³

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