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

The inverse Laplace transform of s/(s+1) is equal to δ(t)-e-t where δ is the Dirac delta function. The inverse Laplace of s/(s+1) is denoted by L-1{s/(s+1)}, and its formula is given by

L-1{s/(s+1)} = δ(t)-e-t.

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

Question: Find the inverse Laplace of s/(s+1).

Solution:

Using the partial fraction decomposition, we have the following:

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

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

= 1 – $\dfrac{1}{s+1}$

So $\dfrac{s}{s+1}$ = 1 -$\dfrac{1}{s+1}$.

Taking inverse Laplace on both sides,

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

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

= δ(t)-e-t as the inverse Laplace of 1 is δ(t) and L-1{1/(s+a)}=e-at.

So the inverse Laplace transform of s/(s+1) is δ(t)-e-t where δ denotes the Dirac delta function.

More Inverse Laplace Transforms:

## FAQs

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

Answer: The inverse Laplace transform of s/s+1 is equal to δ(t)-e-t where δ is the Dirac delta function.

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