---
title: "Inverse Laplace transforms 1/(s²(s²+1))and"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-1-s2-s2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s2-s2
site: BCT Engineering
---

# Inverse Laplace transforms 1/(s²(s²+1))and

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2082 Baishakh, 2+3 marks
- Video: [Inverse Laplace Transform of 1/s²(s²+1) | Convolution Theorem](https://www.youtube.com/watch?v=EV8F1tkbF-U) — RK Eduworld (Classic)
- 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-s2-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of a) 1/(s²(s² + 1)) and b) log(1 + 1/s²).

## Short answer

Partial fractions: 1/(s²(s²+1))=1/s²−1/(s²+1), so L⁻¹ = t − sin t. For log(1+1/s²), use derivative property: L⁻¹ = 2(1−cos t)/t.

## Answer

#### a) L⁻¹{1/(s²(s² + 1))}
Partial fractions: 1/(s²(s² + 1)) = 1/s² − 1/(s² + 1).

```text
L⁻¹ = t − sin t
```

#### b) L⁻¹{log(1 + 1/s²)}
Let F(s) = ln((s² + 1)/s²) = ln(s² + 1) − 2 ln s. Using L{t f(t)} = −F'(s):

```text
F'(s) = 2s/(s² + 1) − 2/s
L{t f(t)} = 2/s − 2s/(s² + 1)
t f(t) = 2 − 2 cos t
f(t) = 2(1 − cos t)/t
```

**Answers:** a) t − sin t; b) 2(1 − cos t)/t.

## Check yourself

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

1. What is L^-1{1/(s^2 (s^2+1))}?
   A. t + sin t
   B. 1 - cos t
   C. t - cos t
   D. t - sin t
   Answer: D. 1/(s^2(s^2+1)) = 1/s^2 - 1/(s^2+1). (Key point: Part (a))

2. For F(s) = ln(s^2+1) - 2 ln s, what is F'(s)?
   A. 2s/(s^2+1) - 2/s
   B. 2s/(s^2+1) + 2/s
   C. 1/(s^2+1) - 1/s
   D. s/(s^2+1) - 1/s
   Answer: A. Differentiate each log: (2s)/(s^2+1) and 2/s. (Key point: Derivative of F)

3. What is t f(t) = L^-1{-F'(s)}?
   A. 2 + 2 cos t
   B. 2 - 2 cos t
   C. 2 cos t - 2
   D. 2 sin t
   Answer: B. -F' = 2/s - 2s/(s^2+1), which inverts to 2 - 2 cos t. (Key point: Step before dividing by t)

4. What is L^-1{log(1 + 1/s^2)}?
   A. 2(1 - cos t)
   B. (1 - cos t)/t
   C. 2(1 - cos t)/t
   D. 2(1 + cos t)/t
   Answer: C. Divide t f(t) = 2 - 2 cos t by t. (Key point: Result for part (b))

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