---
title: "Inverse Laplace transforms s/((s−3)(s²+4))"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-s-s-3"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s-s-3
site: BCT Engineering
---

# Inverse Laplace transforms s/((s−3)(s²+4))

Mathematics III · Chapter 3: Laplace Transform

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

## Question

Find the inverse Laplace transform of a) s/((s − 3)(s² + 4))  b) log(s(s + 1)/(s² + 4)).

## Short answer

For (a), partial fractions give (1/13)[3/(s−3) −3s/(s²+4) +4/(s²+4)], inverse is (1/13)(3e^(3t) −3 cos 2t +2 sin 2t). For (b), differentiate F(s), use L[t f(t)] = −F′(s) to get f(t) = (2 cos 2t −1 − e^(−t))/t.

## Answer

#### a) L⁻¹[s/((s − 3)(s² + 4))]
```text
s/((s−3)(s²+4)) = A/(s−3) + (Bs + C)/(s²+4)
s = A(s² + 4) + (Bs + C)(s − 3)
s = 3:  3 = 13A → A = 3/13
s²: 0 = A + B → B = −3/13
const: 0 = 4A − 3C → C = 4A/3 = 4/13
```

So F(s) = (1/13) [ 3/(s − 3) − 3s/(s² + 4) + 4/(s² + 4) ].

**Answer: (1/13) [ 3e^(3t) − 3 cos 2t + 2 sin 2t ]**

#### b) L⁻¹[log(s(s + 1)/(s² + 4))]
F(s) = ln s + ln(s + 1) − ln(s² + 4).
F′(s) = 1/s + 1/(s + 1) − 2s/(s² + 4).
L[t f(t)] = −F′(s) = −1/s − 1/(s + 1) + 2s/(s² + 4):

```text
t f(t) = −1 − e^(−t) + 2 cos 2t
```

**Answer: f(t) = (2 cos 2t − 1 − e^(−t))/t**

## Check yourself

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

1. Partial fractions for s/((s-3)(s^2+4)) use
   A. A/(s-3) + (Bs+C)/(s^2+4)
   B. A/(s-3) + B/(s^2+4)
   C. A/(s-3) + B/(s+2) + C/(s-2)
   D. A/(s-3) + Bs/(s^2+4)
   Answer: A. An irreducible quadratic needs a linear numerator. (Key point: Form with quadratic factor)

2. A equals
   A. 1/13
   B. 3/13
   C. 3/4
   D. -3/13
   Answer: B. At s = 3: 3 = A(9+4). (Key point: Constant A)

3. L^-1[s/((s-3)(s^2+4))] =
   A. (1/13)[3e^(3t) - 3 cos 2t + 4 sin 2t]
   B. (1/13)[3e^(3t) + 3 cos 2t + 2 sin 2t]
   C. (1/13)[3e^(3t) - 3 cos 2t + 2 sin 2t]
   D. (1/13)[3e^(-3t) - 3 cos 2t + 2 sin 2t]
   Answer: C. 4/(s^2+4) inverts to 2 sin 2t. (Key point: Result (a))

4. L^-1[log(s(s+1)/(s^2+4))] =
   A. (1 + e^(-t) - 2 cos 2t)/t
   B. 2 cos 2t - 1 - e^(-t)
   C. (cos 2t - 1 - e^(-t))/t
   D. (2 cos 2t - 1 - e^(-t))/t
   Answer: D. -F'(s) = -1/s - 1/(s+1) + 2s/(s^2+4). (Key point: Result (b))

## Close variants set in other exams

- [Inverse Laplace transforms 1/(s²(s²+1))and](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-1-s2-s2) — Mathematics III, chapter 3, asked 14×
- [Inverse Laplace transforms log((s+1)/(s−1))](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-log-s-1) — Mathematics III, chapter 3, asked 14×
- [Inverse Laplace transforms log(s/(s+1))](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-log-s-s) — Mathematics III, chapter 3, asked 14×
- [Inverse Laplace transforms (s+2)/(s²−4s+13)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-s-2-s2) — Mathematics III, chapter 3, asked 14×
- [Inverse Laplace transforms 1/((s−2)(s+4))](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-1-s-2-three) — Mathematics III, chapter 3, asked 14×
- [Inverse Laplace transforms 1/(s²(s²+a²))](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-1-s2-s2-two) — Mathematics III, chapter 3, asked 14×
- [Inverse Laplace transforms 1/(s(s−1))](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-1-s-s) — Mathematics III, chapter 3, asked 23×
- [Inverse Laplace transforms s²/(s−1)³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-s2-s-1) — Mathematics III, chapter 3, asked 23×
- [Inverse Laplace transforms (s+3)/(s²+6s+13)²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-s-3-s2) — Mathematics III, chapter 3, asked 23×
- [Inverse Laplace transforms 1/(s²−3s+2)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-1-s2-3s) — Mathematics III, chapter 3, asked 23×
- [Inverse Laplace transforms 1/(s(s+1))](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-1-s-s-two) — Mathematics III, chapter 3, asked 23×
- [Inverse Laplace transforms (2s+3)/(s²+5s−6)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-2s-3-s2) — Mathematics III, chapter 3, asked 23×
- [Inverse Laplace transforms (s+2)³/s⁴](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-laplace-transforms-s-2-3) — Mathematics III, chapter 3, asked 23×