---
title: "Integral of x/(1+x²)² from 1 to ∞"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-1-to-infinity-x-over-1-plus-x2-squared"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-1-to-infinity-x-over-1-plus-x2-squared
site: BCT Engineering
---

# Integral of x/(1+x²)² from 1 to ∞

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2067 Asoj
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-1-to-infinity-x-over-1-plus-x2-squared
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Evaluate ∫₁^∞ x dx/(1+x²)².

## Short answer

Substitute t=1+x², dt=2x dx. The integral becomes (1/2)∫_{2}^{∞} t^{-2} dt = (1/2)[−1/t]_{2}^{∞} = (1/2)(0+1/2) = 1/4.

## Answer

Let I = ∫₁^∞ x dx/(1 + x²)². This is an improper integral (infinite upper limit), defined as lim_(b→∞) ∫₁^b.

Put t = 1 + x², dt = 2x dx. Limits: x = 1 → t = 2; x = b → t = 1 + b².

```text
I = lim_(b→∞) (1/2) ∫₂^(1+b²) t⁻² dt
  = lim (1/2) [−1/t]₂^(1+b²)
  = lim (1/2) [1/2 − 1/(1 + b²)]
  = (1/2)(1/2 − 0)
```
**Answer: I = 1/4** (the integral converges).

## Check yourself

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

1. The improper integral is defined as
   A. limit of the integral from 1 to b as b → ∞
   B. the integral from 1 to 0
   C. the integral at x = ∞ directly
   D. the sum of values at integers
   Answer: A. Infinite limits are handled via limits. (Key point: Definition)

2. Put t = 1 + x^2: the integral becomes
   A. integral t^-2 dt from 1 to b
   B. (1/2) integral t^-2 dt from 2 to 1+b^2
   C. 2 integral t^-2 dt from 2 to b
   D. (1/2) integral t^-1 dt
   Answer: B. dt = 2x dx. (Key point: Substitution)

3. I equals
   A. 1/2
   B. 1/8
   C. 1/4 (converges)
   D. diverges
   Answer: C. (1/2)(1/2 - 0) = 1/4. (Key point: Result)

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