---
title: "Gauss 3-point: 1/(x²+2x+1)"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:gauss-3pt-1-over-x2-2x-1"
source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-3pt-1-over-x2-2x-1
site: BCT Engineering
---

# Gauss 3-point: 1/(x²+2x+1)

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2071 Bhadra, 5 marks
- Video: [Evaluate the Integral by using Gauss Quadrature Three Point Formula (ME 8692 FEA UNIT V)](https://www.youtube.com/watch?v=7JYloRa9LrE) — shanmuga Ramanan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-3pt-1-over-x2-2x-1
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Evaluate ∫ from 0 to 2 of dx/(x² + 2x + 1) using Gaussian 3 point formula.

## Short answer

Map [0,2] to [-1,1] with x = t+1, dx = dt. At Gauss nodes t = -0.7746, 0, 0.7746, f(x)=1/(x+1)² gives 0.6660, 0.2500, 0.1299; weighted sum 0.6644. Integral ≈ 0.6644.

## Answer

The 3-point Gauss–Legendre formula works on [−1, 1]:

∫₋₁¹ g(t) dt ≈ (5/9)·g(−√0.6) + (8/9)·g(0) + (5/9)·g(√0.6),   √0.6 = 0.774597

#### Change of limits

For ∫ₐᵇ f(x) dx put x = (b−a)/2·t + (b+a)/2. Here a = 0, b = 2, so

```text
x = 1.000000·t + 1.000000,     dx = 1.000000 dt
∫ f(x) dx = 1.000000 ∫₋₁¹ f(1.000000t + 1.000000) dt
```

#### Nodes and function values

| t | x = 1.0000t + 1.0000 | f(x) | weight w | w·f(x) |
|---|---|---|---|---|
| -0.774597 | 0.225403 | 0.665950 | 5/9 | 0.369972 |
| 0.000000 | 1.000000 | 0.250000 | 8/9 | 0.222222 |
| 0.774597 | 1.774597 | 0.129897 | 5/9 | 0.072165 |

```text
I ≈ 1.000000 × [ 0.664360 ] = 0.6644
```

(The exact value is ∫₀² dx/(x+1)² = 1 − 1/3 = 0.6667, so the 3-point result is accurate to about two decimal places.)

**Answer: I ≈ 0.6644**

## Check yourself

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

1. What are the nodes and weights of the 3-point Gauss-Legendre formula on [-1, 1]?
   A. Nodes -0.774597, 0, 0.774597 with weights 8/9, 5/9, 8/9
   B. Nodes -0.577350, 0.577350 with weights 1, 1
   C. Nodes -0.774597, 0, 0.774597 with weights 5/9, 8/9, 5/9
   D. Nodes -1, 0, 1 with weights 1/3, 4/3, 1/3
   Answer: C. The 3-point rule uses t = 0, ±sqrt(0.6) with weight 8/9 at the centre and 5/9 at each outer node. (Key point: Gauss 3-point weights)

2. How is the integral over [0, 2] converted to [-1, 1] here?
   A. x = 2t + 1, dx = 2 dt
   B. x = t + 2, dx = dt
   C. x = 0.5t + 1, dx = 0.5 dt
   D. x = t + 1, dx = dt
   Answer: D. (b-a)/2 = 1 and (b+a)/2 = 1, so the factor in front is 1. (Key point: Change of limits)

3. After computing w·f(x) at the three nodes, what is the last step?
   A. Add the three products and multiply by the factor (b-a)/2 from dx
   B. Add the three products and divide by 3
   C. Average the three f(x) values and multiply by b-a
   D. Add the products and multiply by 2/(b-a)
   Answer: A. Since dx = ((b-a)/2) dt, the weighted sum is multiplied by (b-a)/2. (Key point: Final assembly)

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