---
title: "Contour integral ∫x²/((1+x²)(x²+4))"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:real-integral-x2-1-x2-x2-4"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/real-integral-x2-1-x2-x2-4
site: BCT Engineering
---

# Contour integral ∫x²/((1+x²)(x²+4))

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2075 Baisakh, 5 marks
- Video: [Evaluation of Improper Integrals by Contour Integration](https://www.youtube.com/watch?v=Vb_g0lsPDq8) — Maths by Nidhi
- 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-x2-1-x2-x2-4
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate the integral by contour integration: ∫_{-∞}^{∞} x² / ((1 + x²)(x² + 4)) dx.

## Short answer

Use contour integration with f(z)=z²/((z²+1)(z²+4)). Poles in upper half-plane: z=i and z=2i. Residues sum to 1/(6i); upper semicircle vanishes. Result: π/3.

## Answer

**Method.** Let I = ∫_{−∞}^{∞} x²/((1 + x²)(x² + 4)) dx. Consider f(z) = z²/((z² + 1)(z² + 4)) around the closed contour C made of the real segment [−R, R] and the upper semicircle Γ: |z| = R. 

Poles of f: z = ±i and z = ±2i (all simple). Those in the upper half-plane (inside C for R > 2): **z = i and z = 2i**.

**Residues:**

- At z = i: Res = lim (z − i) f(z) = i²/((2i)(−1 + 4)) = −1/(6i)
- At z = 2i: Res = (2i)²/(((2i)² + 1)(4i)) = −4/((−3)(4i)) = 1/(3i)

Sum = −1/(6i) + 1/(3i) = 1/(6i).

**Residue theorem:** ∮_C f dz = 2πi · 1/(6i) = π/3.

**Semicircle vanishes:** on Γ, |f(z)| ≤ R²/((R² − 1)(R² − 4)) ~ 1/R², so |∫_Γ f dz| ≤ (π R)(1/R²) → 0 as R → ∞.

Hence ∫_{−∞}^{∞} f(x) dx = π/3.

**Answer: I = π/3**

(Check by partial fractions: x²/((x² + 1)(x² + 4)) = (−1/3)/(x² + 1) + (4/3)/(x² + 4); integral = −π/3 + (4/3)(π/2) = π/3.)

## Check yourself

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

1. What contour is used and which poles are enclosed?
   A. Full circle; all four poles
   B. Upper semicircle; poles -i and -2i
   C. Real segment plus upper semicircle; poles z = i and 2i
   D. Lower semicircle; poles i and 2i
   Answer: C. Upper half-plane poles are i and 2i. (Key point: Contour)

2. Why does the semicircle integral vanish as R→∞?
   A. f is analytic on the arc
   B. The arc length tends to 0
   C. |f| ~ R so it cancels
   D. |f| ≲ 1/R² so the integral is bounded by πR/R² → 0
   Answer: D. ML-inequality: length πR times 1/R². (Key point: Arc vanishes)

3. What is the sum of residues at i and 2i?
   A. 1/(6i)
   B. 1/(2i)
   C. -1/(6i)
   D. 1/(3i)
   Answer: A. -1/(6i) + 1/(3i) = 1/(6i). (Key point: Residue sum)

4. What is the value of the integral?
   A. π/6
   B. π/3
   C. 2π/3
   D. π
   Answer: B. 2πi × 1/(6i) = π/3. (Key point: Result)

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