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}.

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## 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:

- Table of Inverse Laplace Transformations
- Inverse Laplace transform of 1
- Inverse Laplace transform of 1/s
- Find L
^{-1}{1/s^{2}} - Inverse Laplace transform of 1/s
^{3} - Find L
^{-1}{1/s(s+1)}

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