---
title: "Inverse Laplace transforms 1/((s−2)(s+4))"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-1-s-2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s-2
site: BCT Engineering
---

# Inverse Laplace transforms 1/((s−2)(s+4))

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2083 Baishakh, 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-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of (i) 1/((s − 2)(s + 4))  (ii) cot⁻¹(s + 1).

## Short answer

Partial fractions give L⁻¹{1/((s−2)(s+4))}=(e^(2t)−e^(−4t))/6. For cot⁻¹(s+1), use L{t f(t)}=−F'(s) to obtain f(t)=e^(−t) sin t / t.

## Answer

#### (i) L⁻¹{1/((s − 2)(s + 4))}
Partial fractions: 1/((s − 2)(s + 4)) = A/(s − 2) + B/(s + 4), with A = 1/6 (put s = 2), B = −1/6 (put s = −4).

```text
L⁻¹ = (1/6) e^(2t) − (1/6) e^(−4t) = (e^(2t) − e^(−4t))/6
```

#### (ii) L⁻¹{cot⁻¹(s + 1)}
Let F(s) = cot⁻¹(s + 1) and f(t) = L⁻¹{F}. Using L{t f(t)} = −F'(s):

```text
F'(s) = −1/(1 + (s + 1)²)
L{t f(t)} = 1/((s + 1)² + 1)
t f(t) = e^(−t) sin t       (first shifting theorem)
f(t) = e^(−t) sin t / t
```

**Answers:** (i) (e^(2t) − e^(−4t))/6; (ii) e^(−t) (sin t)/t.

## Check yourself

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

1. 1/((s-2)(s+4)) = A/(s-2) + B/(s+4). What is A?
   A. 1/4
   B. -1/6
   C. 6
   D. 1/6
   Answer: D. Put s = 2 in the other factor: A = 1/(2+4) = 1/6. (Key point: Coefficient A)

2. What is L^-1{1/((s-2)(s+4))}?
   A. (e^(2t) - e^(-4t))/6
   B. (e^(-2t) - e^(4t))/6
   C. (e^(2t) + e^(-4t))/6
   D. e^(2t) - e^(-4t)
   Answer: A. With A = 1/6 and B = -1/6, the inverse is (1/6)e^(2t) - (1/6)e^(-4t). (Key point: Inverse of part (i))

3. Which relation is used to invert cot^-1(s+1)?
   A. Partial fractions on cot^-1(s+1)
   B. L{t f(t)} = -F'(s): differentiate F, invert, divide by t
   C. Convolution of two simpler functions
   D. Second shifting theorem
   Answer: B. Inverse trig functions of s are handled by differentiating F, which gives a rational function that is easy to invert. (Key point: Method for cot^-1(s+1))

4. What is F'(s) for F(s) = cot^-1(s+1)?
   A. 1/((s+1)^2 + 1)
   B. -1/(s^2 + 1)
   C. -1/((s+1)^2 + 1)
   D. -1/(1 + (s-1)^2)
   Answer: C. d/dx cot^-1 x = -1/(1+x^2) with x = s+1. (Key point: Derivative of F)

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