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:

• Table of Inverse Laplace Transformations
• Inverse Laplace transform of 1
• Inverse Laplace transform of 1/s
• Find L^-1{1/s²}
• Inverse Laplace transform of 1/s³
• Find L^-1{1/s(s+1)}

FAQs