---
title: "Inverse Laplace transforms 1/(s²(s²+a²))"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-1-s2-s2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s2-s2-two
site: BCT Engineering
---

# Inverse Laplace transforms 1/(s²(s²+a²))

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2067 Asar
- 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of: a) 1/(s²(s² + a²))  b) log((s + 1)/(s − 1)).

## Short answer

For (a), partial fractions give (1/a²)[1/s² − 1/(s²+a²)], inverse is (at − sin at)/a³. For (b), differentiate F(s) to get F′(s), then L[t f(t)] = −F′(s) yields f(t) = 2 sinh t / t.

## Answer

#### a) L⁻¹[1/(s²(s² + a²))]
```text
1/(s²(s²+a²)) = (1/a²) [ 1/s² − 1/(s² + a²) ]
```

Check: 1/s² − 1/(s² + a²) = a²/(s²(s² + a²)) ✓.

**Answer: (1/a²) [ t − (sin at)/a ] = (at − sin at)/a³**

#### b) L⁻¹[log((s + 1)/(s − 1))]
F(s) = ln(s + 1) − ln(s − 1), F′(s) = 1/(s + 1) − 1/(s − 1).
Since L[t f(t)] = −F′(s) = 1/(s − 1) − 1/(s + 1):

```text
t f(t) = eᵗ − e^(−t) = 2 sinh t
```

**Answer: f(t) = 2 sinh t / t**

## Check yourself

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

1. 1/(s^2(s^2+a^2)) can be written as
   A. (1/a^2)[1/s^2 - 1/(s^2+a^2)]
   B. (1/a^2)[1/s^2 + 1/(s^2+a^2)]
   C. a^2[1/s^2 - 1/(s^2+a^2)]
   D. (1/a^2)[1/s - 1/(s^2+a^2)]
   Answer: A. The difference of the two terms has numerator a^2. (Key point: Partial fractions)

2. L^-1[1/(s^2(s^2+a^2))] equals
   A. (at + sin at)/a^3
   B. (at - sin at)/a^3
   C. (t - sin at)/a^2
   D. (at - sin at)/a^2
   Answer: B. L^-1[1/s^2] = t and L^-1[1/(s^2+a^2)] = sin(at)/a. (Key point: Result (a))

3. For F(s) = ln(s+1) - ln(s-1), -F'(s) equals
   A. 1/(s+1) - 1/(s-1)
   B. 1/(s-1) + 1/(s+1)
   C. 1/(s-1) - 1/(s+1)
   D. ln(s-1) + ln(s+1)
   Answer: C. F' = 1/(s+1) - 1/(s-1). (Key point: Log derivative)

4. L^-1[log((s+1)/(s-1))] equals
   A. 2 cosh t / t
   B. 2 sinh t
   C. sinh t / t
   D. 2 sinh t / t
   Answer: D. t f(t) = e^t - e^(-t) = 2 sinh t. (Key point: Result (b))

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