---
title: "DUIS: ∫ log(1+a²x²)/(1+b²x²) dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:duis-log-1-plus-a2x2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-log-1-plus-a2x2
site: BCT Engineering
---

# DUIS: ∫ log(1+a²x²)/(1+b²x²) dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2070 Chaitra; 2067 Asoj (asked 2 times)
- Video: [Evaluate ∫(log(1+a²x²))/(1+b²x²)dx between the limit 0 to infinity](https://www.youtube.com/watch?v=wuxLBXcFOu8) — vikas saxena maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-log-1-plus-a2x2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Apply the method of differentiation under integral sign to evaluate ∫₀^∞ log(1+a²x²)/(1+b²x²) dx.

## Short answer

Differentiate with respect to a, apply partial fractions, and integrate using ∫₀^∞ dx/(1+k²x²)=π/(2k) to get dI/da = π/(b(a+b)). Integrating back with I(0)=0 yields (π/b) log(1 + a/b).

## Answer

Let I(a) = ∫₀^∞ log(1 + a²x²)/(1 + b²x²) dx, with a, b > 0. Differentiate with respect to a:

dI/da = ∫₀^∞ 2a x² / [(1 + a²x²)(1 + b²x²)] dx

**Partial fractions.** Since 1/(1 + a²x²) − 1/(1 + b²x²) = (b² − a²)x² / [(1 + a²x²)(1 + b²x²)],

x² / [(1 + a²x²)(1 + b²x²)] = [1/(b² − a²)] [1/(1 + a²x²) − 1/(1 + b²x²)].

Using ∫₀^∞ dx/(1 + k²x²) = π/(2k):

```text
dI/da = 2a/(b²−a²) · (π/2)(1/a − 1/b)
      = 2a/(b²−a²) · π(b−a)/(2ab)
      = π / (b(a+b))
```
Integrate with respect to a:

I(a) = (π/b) log(a + b) + C

When a = 0 the integrand is log 1 = 0, so I(0) = 0: C = −(π/b) log b.

**Answer: ∫₀^∞ log(1 + a²x²)/(1 + b²x²) dx = (π/b) log((a + b)/b) = (π/b) log(1 + a/b).**

## Check yourself

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

1. dI/da of the integral is
   A. integral 2a/(1+a^2x^2) dx
   B. integral 2ax/[(1+a^2x^2)(1+b^2x^2)] dx
   C. integral 2a x^2 /[(1+a^2x^2)(1+b^2x^2)] dx
   D. integral a x^2/(1+b^2x^2) dx
   Answer: C. d/da log(1+a^2x^2) = 2ax^2/(1+a^2x^2). (Key point: Derivative)

2. The integral of dx/(1+k^2x^2) from 0 to ∞ is
   A. π/k
   B. π k/2
   C. 2/π k
   D. π/(2k)
   Answer: D. It equals (1/k) tan^-1(∞). (Key point: Standard integral)

3. After partial fractions, dI/da equals
   A. π/(b(a+b))
   B. π/(a+b)
   C. π/(ab)
   D. π/(b(a-b))
   Answer: A. 2a/(b^2-a^2) · π(b-a)/(2ab) = π/(b(a+b)). (Key point: dI/da)

4. The integral equals
   A. (π/b) log(a+b)
   B. (π/b) log(1 + a/b)
   C. (π/a) log(1+a/b)
   D. π log(1+a/b)
   Answer: B. With I(0) = 0, C = -(π/b) log b. (Key point: Result)

## Close variants set in other exams

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