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

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

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2078 Kartik, 2.5+2.5 marks
- Video: [Inverse Laplace transform of log and cot^-1 function](https://www.youtube.com/watch?v=OsfEg-jR0Xc) — easy study plus
- 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-log-s-s
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

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

## Short answer

a) L^{-1}{ln(s/(s+1))} = (e^{-t} - 1)/t, obtained via derivative property. b) Partial fractions give L^{-1}{1/((s-2)(s^2+1))} = (e^{2t} - cos t - 2 sin t)/5.

## Answer

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

```text
F'(s) = 1/s − 1/(s + 1)
L{t f(t)} = 1/(s + 1) − 1/s  → t f(t) = e^(−t) − 1
f(t) = (e^(−t) − 1)/t
```

#### b) L⁻¹{1/((s − 2)(s² + 1))}
Write 1/((s − 2)(s² + 1)) = A/(s − 2) + (Bs + C)/(s² + 1), so 1 = A(s² + 1) + (Bs + C)(s − 2).

```text
s = 2:   1 = 5A → A = 1/5
s²:      A + B = 0 → B = −1/5
const:   A − 2C = 1 → C = −2/5
F(s) = (1/5)/(s − 2) − (1/5)(s + 2)/(s² + 1)
f(t) = (1/5)[e^(2t) − cos t − 2 sin t]
```

**Answers:** a) (e^(−t) − 1)/t; b) (e^(2t) − cos t − 2 sin t)/5.

## Check yourself

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

1. For F = ln s - ln(s+1), what is L{t f(t)}?
   A. 1/s - 1/(s+1)
   B. ln(s/(s+1))
   C. 1/(s(s+1))
   D. 1/(s+1) - 1/s
   Answer: D. -F' = -(1/s - 1/(s+1)). (Key point: Relation used)

2. What is L^-1{log(s/(s+1))}?
   A. (e^(-t) - 1)/t
   B. (1 - e^(-t))/t
   C. e^(-t) - 1
   D. (e^t - 1)/t
   Answer: A. t f = e^(-t) - 1, so f = (e^(-t) - 1)/t. (Key point: Result for (a))

3. 1 = A(s^2+1) + (Bs+C)(s-2). With A = 1/5, constants give A - 2C = 1. C = ?
   A. 2/5
   B. -2/5
   C. -1/5
   D. 1/10
   Answer: B. 2C = A - 1 = -4/5, so C = -2/5. (Key point: Finding C in (b))

4. What is L^-1{1/((s-2)(s^2+1))}?
   A. (e^(2t) + cos t + 2 sin t)/5
   B. (e^(2t) - cos t + 2 sin t)/5
   C. (e^(2t) - cos t - 2 sin t)/5
   D. (e^(2t) - 2 cos t - sin t)/5
   Answer: C. -(1/5)(s+2)/(s^2+1) inverts to -(cos t + 2 sin t)/5. (Key point: Result for (b))

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