---
title: "Inverse Laplace transforms (s+2)/(s+1)⁴"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-s-2-s"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s-2-s
site: BCT Engineering
---

# Inverse Laplace transforms (s+2)/(s+1)⁴

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2075 Chaitra, 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-s-2-s
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of: i) (s + 2)/(s + 1)⁴  ii) cot⁻¹(s + 1).

## Short answer

i) (s+2)/(s+1)⁴ = 1/(s+1)³ + 1/(s+1)⁴; inverse is e^(−t)(t²/2 + t³/6). ii) Differentiate cot⁻¹(s+1) to −1/(( s+1)²+1); L{t f(t)} = −F'(s) gives t f(t)=e^(−t) sin t, so f(t)=e^(−t) sin t / t.

## Answer

#### i) 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
```

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

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

#### ii) L⁻¹{cot⁻¹(s + 1)}
Let F(s) = cot⁻¹(s + 1). Then F'(s) = −1/((s + 1)² + 1), and with L{t f(t)} = −F'(s):

```text
L{t f(t)} = 1/((s + 1)² + 1)  → t f(t) = e^(−t) sin t
f(t) = e^(−t) sin t / t
```

**Answers:** i) e^(−t)(t²/2 + t³/6); ii) e^(−t)(sin t)/t.

## Check yourself

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

1. (s+2)/(s+1)^4 becomes what?
   A. 1/(s+1)^2 + 1/(s+1)^4
   B. 2/(s+1)^3
   C. 1/(s+1)^3 + 1/(s+1)^4
   D. 1/(s+1)^3 - 1/(s+1)^4
   Answer: C. s + 2 = (s+1) + 1. (Key point: Split)

2. What is the inverse?
   A. e^(-t)(t^2/2 + t^3/3)
   B. e^(t)(t^2/2 + t^3/6)
   C. e^(-t)(t + t^2/2)
   D. e^(-t)(t^2/2 + t^3/6)
   Answer: D. Using e^(-t) t^(n-1)/(n-1)!. (Key point: Result for (i))

3. What is L^-1{cot^-1(s+1)}?
   A. e^(-t) sin t / t
   B. e^(-t) sin t
   C. e^(t) sin t / t
   D. e^(-t) cos t / t
   Answer: A. -F' = 1/((s+1)^2+1), t f = e^(-t) sin t. (Key point: Result for (ii))

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