---
title: "Show ∫ log(1+x²)/(1+x²) dx = π log 2"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-log-1-plus-x2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-log-1-plus-x2
site: BCT Engineering
---

# Show ∫ log(1+x²)/(1+x²) dx = π log 2

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2073 Chaitra; 2070 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-log-1-plus-x2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Show that ∫₀^∞ log(1+x²)/(1+x²) dx = π log 2.

## Short answer

Put x = tanθ, then I = -2∫₀^{π/2} log cosθ dθ. Using the known result ∫₀^{π/2} log cosθ dθ = −(π/2) log 2, we find I = π log 2.

## Answer

Let I = ∫₀^∞ log(1 + x²)/(1 + x²) dx.

Put x = tan θ, dx = sec²θ dθ, 1 + x² = sec²θ. Limits: 0 → 0, ∞ → π/2.

I = ∫₀^(π/2) log(sec²θ) dθ = −2 ∫₀^(π/2) log cos θ dθ.

**Evaluate J = ∫₀^(π/2) log cos θ dθ.** By the property ∫₀^a f(θ)dθ = ∫₀^a f(a − θ)dθ, J = ∫₀^(π/2) log sin θ dθ. Add:

```text
2J = ∫₀^(π/2) log(sinθ cosθ) dθ
   = ∫₀^(π/2) [log sin 2θ − log 2] dθ
   = ∫₀^(π/2) log sin 2θ dθ − (π/2) log 2
```
Put 2θ = u: ∫₀^(π/2) log sin 2θ dθ = (1/2)∫₀^π log sin u du = ∫₀^(π/2) log sin u du = J. So 2J = J − (π/2) log 2, giving

J = −(π/2) log 2.

I = −2J = −2 · (−(π/2) log 2) = π log 2.

**Proved: ∫₀^∞ log(1 + x²)/(1 + x²) dx = π log 2.**

## Check yourself

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

1. Putting x = tan θ turns the integral into
   A. 2 integral of log sin θ dθ from 0 to π
   B. integral of log cos θ dθ from 0 to ∞
   C. -2 integral of log cos θ dθ from 0 to π/2
   D. -integral of log tan θ dθ from 0 to π/2
   Answer: C. log(sec^2 θ) = -2 log cos θ. (Key point: Substitution)

2. J (log cos) equals the integral of log sin because
   A. cos and sin are equal
   B. of integration by parts
   C. sin θ = cos θ on the interval
   D. of the property f(θ) → f(a-θ) with a = π/2
   Answer: D. Replacing θ by π/2 - θ swaps sin and cos. (Key point: Key trick)

3. Adding gives 2J = J - (π/2) log 2, so J equals
   A. -(π/2) log 2
   B. (π/2) log 2
   C. -π log 2
   D. 0
   Answer: A. Solving 2J = J - (π/2) log 2 gives J = -(π/2) log 2. (Key point: Value of J)

4. The integral equals
   A. (π/2) log 2
   B. π log 2
   C. -π log 2
   D. π^2 log 2
   Answer: B. I = -2J = π log 2. (Key point: Result)

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