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

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

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2071 Chaitra, 2.5+2.5 marks
- 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-two
- 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) s²/(s² + b²)².

## Short answer

a) 1/(s(s+1)) = 1-e^{-t}. b) Convolution with cos bt gives (1/(2b))(sin bt + bt cos bt).

## Answer

#### a) L⁻¹[1/(s(s + 1))]
1/(s(s+1)) = 1/s − 1/(s+1).

**Answer: 1 − e^(−t)**

#### b) L⁻¹[s²/(s² + b²)²]
Use the convolution theorem with F = G = s/(s² + b²), so f = g = cos bt.

```text
L⁻¹ = ∫₀ᵗ cos bu cos b(t − u) du
    = ½ ∫₀ᵗ [cos bt + cos b(2u − t)] du
    = ½ [ t cos bt + (sin bt)/b ]
```

(The second integral: ∫₀ᵗ cos b(2u − t) du = [sin b(2u − t)/(2b)]₀ᵗ = (sin bt)/b.)

**Answer: (1/2b)(sin bt + bt cos bt)**

## 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. e^(-t) - 1
   B. 1 - e^(-t)
   C. 1 + e^(-t)
   D. e^(-t)
   Answer: B. 1/s - 1/(s+1). (Key point: Result for (a))

2. For s^2/(s^2+b^2)^2, F and G are?
   A. F = G = 1/(s^2+b^2)
   B. F = s, G = 1/(s^2+b^2)^2
   C. F = G = s/(s^2+b^2), f = g = cos bt
   D. f = g = sin bt
   Answer: C. Product of two equal factors. (Key point: Convolution setup)

3. cos bu cos b(t-u) = ?
   A. (1/2)[cos bt - cos b(2u-t)]
   B. cos bt
   C. (1/2)[sin bt + sin b(2u-t)]
   D. (1/2)[cos bt + cos b(2u-t)]
   Answer: D. cosA cosB. (Key point: Product to sum)

4. What is the inverse?
   A. (1/2b)(sin bt + bt cos bt)
   B. (1/2b)(sin bt - bt cos bt)
   C. (1/b)(sin bt + bt cos bt)
   D. (t/2) sin bt
   Answer: A. (1/2)[t cos bt + (sin bt)/b]. (Key point: Result for (b))

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