---
title: "Inverse Laplace transforms 1/((s−2)(s²+1))"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-1-s-2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s-2-two
site: BCT Engineering
---

# Inverse Laplace transforms 1/((s−2)(s²+1))

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2075 Asoj, 2.5+2.5 marks
- Video: [How to Find the Inverse Laplace Transform of 1/((s+1)(s²+2s+2)) in Very Easy Steps #CSIR_NET #GAT...](https://www.youtube.com/watch?v=jLy9-ko7rsk) — EM by danishwar shabir
- 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of: (i) 1/((s − 2)(s² + 1))  (ii) cot⁻¹(s + 1).

## Short answer

i) Partial fractions: 1/(( s−2)(s²+1)) = (1/5)[1/(s−2) − (s+2)/(s²+1)]; inverse is (e^(2t) − cos t − 2 sin t)/5. ii) F(s)=cot⁻¹(s+1), F'(s)=−1/(( s+1)²+1), so t f(t)=e^(−t) sin t → f(t)=e^(−t) sin t / t.

## Answer

#### (i) L⁻¹[1/((s − 2)(s² + 1))]
Partial fractions: 1/((s−2)(s²+1)) = A/(s−2) + (Bs + C)/(s²+1).

```text
1 = A(s² + 1) + (Bs + C)(s − 2)
s = 2:  1 = 5A          → A = 1/5
s² coefficient: 0 = A + B → B = −1/5
constant: 1 = A − 2C    → C = −2/5
```

```text
F(s) = (1/5) [ 1/(s − 2) − (s + 2)/(s² + 1) ]
L⁻¹  = (1/5) [ e^(2t) − cos t − 2 sin t ]
```

**Answer: (e^(2t) − cos t − 2 sin t)/5**

#### (ii) L⁻¹[cot⁻¹(s + 1)]
Let F(s) = cot⁻¹(s + 1). Then F′(s) = −1/((s + 1)² + 1).
Since L[t f(t)] = −F′(s):

```text
L[t f(t)] = 1/((s + 1)² + 1)
t f(t) = e^(−t) sin t
```

**Answer: f(t) = e^(−t) sin t / t**

## Check yourself

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

1. 1 = A(s^2+1) + (Bs+C)(s-2). Put s = 2.
   A. A = 1/2
   B. A = 1/5
   C. A = 1/3
   D. A = 5
   Answer: B. 1 = 5A. (Key point: Coefficient A)

2. What is the inverse?
   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 (i))

3. What is L^-1{cot^-1(s+1)}?
   A. e^(-t) sin t
   B. e^(t) sin t / t
   C. sin t / t
   D. e^(-t) sin t / t
   Answer: D. t f = e^(-t) sin t. (Key point: Result for (ii))

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