---
title: "Inverse Laplace transforms 1/(s(s−1))"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-1-s-s"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s-s
site: BCT Engineering
---

# Inverse Laplace transforms 1/(s(s−1))

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2082 Shrawan, 2+3 marks
- Video: [Inverse Laplace Transform Examples || First Shifting Property of Inverse Laplace Transform ||](https://www.youtube.com/watch?v=IVJjm5FE4x8) — Let's Unlock Maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s-s
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of a) 1/(s(s − 1))  b) (s + 2)/(s + 1)⁴.

## Short answer

Decompose: 1/(s(s−1))=1/(s−1)−1/s ⇒ L⁻¹=eᵕ−1. Write (s+2)/(s+1)⁴=1/(s+1)³+1/(s+1)⁴, invert to e^(−t)(t²/2 + t³/6).

## Answer

#### a) L⁻¹{1/(s(s − 1))}
1/(s(s − 1)) = 1/(s − 1) − 1/s.

```text
L⁻¹ = eᵗ − 1
```

#### b) L⁻¹{(s + 2)/(s + 1)⁴}
Write s + 2 = (s + 1) + 1:

```text
(s + 2)/(s + 1)⁴ = 1/(s + 1)³ + 1/(s + 1)⁴
```

Use L⁻¹{1/(s + 1)ⁿ} = e^(−t) t^(n−1)/(n − 1)!:

```text
L⁻¹ = e^(−t) t²/2! + e^(−t) t³/3! = e^(−t) (t²/2 + t³/6)
```

**Answers:** a) eᵗ − 1; b) e^(−t) (t²/2 + t³/6) = e^(−t) t² (3 + t)/6.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is L^-1{1/(s(s-1))}?
   A. 1 - e^t
   B. e^t + 1
   C. e^(-t) - 1
   D. e^t - 1
   Answer: D. 1/(s(s-1)) = 1/(s-1) - 1/s, which inverts to e^t - 1. (Key point: Part (a))

2. What is the useful first step?
   A. Write s + 2 = (s+1) + 1 and split into 1/(s+1)^3 + 1/(s+1)^4
   B. Apply the second shifting theorem
   C. Cancel (s+1) leaving 1/(s+1)^3
   D. Multiply numerator and denominator by (s+1)
   Answer: A. Rewriting the numerator in terms of (s+1) gives terms of the form 1/(s+1)^n that invert directly. (Key point: First step for (s+2)/(s+1)^4)

3. What is L^-1{1/(s+1)^4}?
   A. e^(-t) t^4/24
   B. e^(-t) t^3/6
   C. e^(-t) t^3/3
   D. e^(t) t^3/6
   Answer: B. L^-1{1/(s+1)^n} = e^(-t) t^(n-1)/(n-1)!, so n = 4 gives t^3/3! = t^3/6. (Key point: Inverse of 1/(s+1)^n)

4. What is L^-1{(s+2)/(s+1)^4}?
   A. e^(-t)(t^2/2 + t^3/3)
   B. e^(-t)(t + t^2/2)
   C. e^(-t)(t^2/2 + t^3/6)
   D. e^(t)(t^2/2 + t^3/6)
   Answer: C. Adding e^(-t)t^2/2! and e^(-t)t^3/3!. (Key point: Result for part (b))

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