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

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

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2073 Chaitra, 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-1-s-s-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

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

## Short answer

For 1/(s(s+2)), partial fractions give ½(1/s -1/(s+2)), so inverse is (1 - e^(-2t))/2. For tan^(-1)(1/s), differentiate: F'(s) = -1/(s^2+1), thus t f(t) = sin t, so f(t) = sin t / t.

## Answer

#### (i) L⁻¹[1/(s(s + 2))]
```text
1/(s(s+2)) = (1/2) [ 1/s − 1/(s + 2) ]
```

**Answer: (1 − e^(−2t))/2**

#### (ii) L⁻¹[tan⁻¹(1/s)]
Let F(s) = tan⁻¹(1/s). Then F′(s) = [1/(1 + 1/s²)](−1/s²) = −1/(s² + 1).
Since L[t f(t)] = −F′(s) = 1/(s² + 1), we have t f(t) = sin t.

**Answer: f(t) = sin t / t**

## Check yourself

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

1. 1/(s(s+2)) =
   A. (1/2)[1/s + 1/(s+2)]
   B. [1/s - 1/(s+2)]
   C. (1/2)[1/(s+2) - 1/s]
   D. (1/2)[1/s - 1/(s+2)]
   Answer: D. Cover-up gives 1/2 and -1/2. (Key point: Partial fractions (i))

2. L^-1 =
   A. (1 - e^(-2t))/2
   B. (1 + e^(-2t))/2
   C. (1 - e^(2t))/2
   D. 1 - e^(-2t)
   Answer: A. Invert each term. (Key point: Result (i))

3. F'(s) for tan^-1(1/s) is
   A. 1/(s^2+1)
   B. -1/(s^2+1)
   C. -1/(s^2-1)
   D. -1/s^2
   Answer: B. Chain rule. (Key point: Derivative (ii))

4. L^-1[tan^-1(1/s)] =
   A. sin t
   B. cos t / t
   C. sin t / t
   D. t sin t
   Answer: C. t f(t) = sin t. (Key point: Result (ii))

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