---
title: "Contour integral ∫x²/(1+x²)³"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:real-integral-x2-over-1-x2-cubed"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/real-integral-x2-over-1-x2-cubed
site: BCT Engineering
---

# Contour integral ∫x²/(1+x²)³

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Bhadra, 5 marks
- Video: [By Using Contour Integration Type 02 Problems 01 -Complex Integration](https://www.youtube.com/watch?v=_CxdH_-Yfks) — SK Maths Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/real-integral-x2-over-1-x2-cubed
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate the real integral ∫_{-∞}^{∞} x²/(1 + x²)³ dx by contour integration in the complex plane.

## Short answer

Take f(z)=z²/(z²+1)³. Only z=i (pole of order 3) inside contour. Residue = –i/16, so integral = 2πi·(–i/16)=π/8, arc integral tends to zero.

## Answer

**Method.** Let I = ∫_{−∞}^{∞} x²/(1 + x²)³ dx. Take f(z) = z²/(z² + 1)³ = z²/((z − i)³(z + i)³) on the closed contour C: segment [−R, R] plus upper semicircle Γ (R > 1).

Only **z = i (pole of order 3)** lies inside C.

**Residue at z = i:**

Res = (1/2!) d²/dz² [ (z − i)³ f(z) ] at z = i = (1/2) g″(i), where g(z) = z² (z + i)⁻³.

g′ = 2z (z + i)⁻³ − 3z² (z + i)⁻⁴

g″ = 2 (z + i)⁻³ − 12 z (z + i)⁻⁴ + 12 z² (z + i)⁻⁵

At z = i, z + i = 2i: (2i)⁻³ = i/8, (2i)⁻⁴ = 1/16, (2i)⁻⁵ = −i/32.

g″(i) = 2(i/8) − 12 i (1/16) + 12(−1)(−i/32)
= i/4 − 3i/4 + 3i/8 = −i/8

Res = (1/2)(−i/8) = −i/16

**Residue theorem:**

∮_C f dz = 2πi (−i/16) = 2π/16 = π/8.

**Arc integral:** |f(z)| ≤ R²/(R² − 1)³ ~ 1/R⁴ on Γ, so the integral over Γ ≤ πR · 1/R⁴ → 0.

**Answer: I = π/8**

(Check with x = tan θ: I = ∫ sin²θ cos²θ dθ over (−π/2, π/2) = π/8.)

## Check yourself

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

1. What is the order of the pole at z = i?
   A. 1
   B. 3
   C. 2
   D. 6
   Answer: B. (z²+1)³ = (z-i)³(z+i)³. (Key point: Pole order)

2. Which formula gives the residue?
   A. d/dz[(z-i)³ f] at i
   B. (1/3!) d³/dz³[...] at i
   C. (1/2!) d²/dz² [(z-i)³ f(z)] at z = i
   D. lim (z-i) f(z)
   Answer: C. For order m: 1/(m-1)! times the (m-1)-th derivative. (Key point: Residue formula)

3. What is the residue?
   A. i/16
   B. -i/8
   C. 1/16
   D. -i/16
   Answer: D. g''(i) = -i/8, halved is -i/16. (Key point: Residue value)

4. What is the integral?
   A. π/8
   B. π/16
   C. π/4
   D. π/2
   Answer: A. 2πi(-i/16) = π/8. (Key point: Result)

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