---
title: "Inverse Laplace transforms s²/(s+2)³"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-s2-s-2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s2-s-2
site: BCT Engineering
---

# Inverse Laplace transforms s²/(s+2)³

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2079 Bhadra, 2+3 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-s2-s-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of a) s²/(s + 2)³  b) tan⁻¹(2/s).

## Short answer

a) Write s^2/(s+2)^3 = 1/(s+2) - 4/(s+2)^2 + 4/(s+2)^3; inverse is e^{-2t}(1-4t+2t^2). b) Using derivative property, L^{-1}{tan^{-1}(2/s)} = (sin 2t)/t.

## Answer

#### a) L⁻¹{s²/(s + 2)³}
Write s² = (s + 2)² − 4(s + 2) + 4:

```text
s²/(s + 2)³ = 1/(s + 2) − 4/(s + 2)² + 4/(s + 2)³
L⁻¹ = e^(−2t) − 4t e^(−2t) + 2t² e^(−2t) = e^(−2t)(1 − 4t + 2t²)
```

#### b) L⁻¹{tan⁻¹(2/s)}
Let F(s) = tan⁻¹(2/s) and f = L⁻¹{F}.

```text
F'(s) = −2/(s² + 4)
L{t f(t)} = −F'(s) = 2/(s² + 4)  → t f(t) = sin 2t
f(t) = sin 2t / t
```

**Answers:** a) e^(−2t)(1 − 4t + 2t²); b) sin 2t / t.

## Check yourself

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

1. Write s^2 in powers of (s+2).
   A. (s+2)^2 - 4(s+2) + 4
   B. (s+2)^2 + 4(s+2) + 4
   C. (s+2)^2 - 4(s+2) - 4
   D. (s+2)^2 - 2(s+2) + 1
   Answer: A. Expanding: s^2 + 4s + 4 - 4s - 8 + 4 = s^2. (Key point: Rewrite s^2)

2. The term 4/(s+2)^3 inverts to?
   A. 4 t^2 e^(-2t)
   B. 2 t^2 e^(-2t)
   C. t^2 e^(-2t)
   D. 2 t e^(-2t)
   Answer: B. L^-1{1/(s+2)^3} = t^2 e^(-2t)/2, times 4. (Key point: Term with 1/(s+2)^3)

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

4. For F = tan^-1(2/s), F' = -2/(s^2+4). What is L^-1{F}?
   A. sin 2t
   B. 2 sin 2t / t
   C. cos 2t / t
   D. sin 2t / t
   Answer: D. t f = L^-1{2/(s^2+4)} = sin 2t, so f = sin 2t / t. (Key point: Part (b))

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