---
title: "DUIS: ∫ tan⁻¹(ax)/(x(1+x²)) dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:duis-atan-ax-over-x-1-plus-x2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-atan-ax-over-x-1-plus-x2
site: BCT Engineering
---

# DUIS: ∫ tan⁻¹(ax)/(x(1+x²)) dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2069 Asar; 2067 Asar (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/duis-atan-ax-over-x-1-plus-x2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Use the method of differentiation under integral sign to evaluate ∫₀^∞ tan⁻¹(ax)/(x(1+x²)) dx.

## Short answer

Differentiate with respect to a, then use partial fractions to find dI/da = π/(2(1+a)). Integrating and using I(0)=0 gives (π/2) log(1+a).

## Answer

**Note:** some copies print the upper limit as α; the intended limit is ∞ (with a finite upper limit the integral has no neat closed form).

Let I(a) = ∫₀^∞ tan⁻¹(ax)/(x(1 + x²)) dx. Differentiate with respect to a:

∂/∂a tan⁻¹(ax) = x/(1 + a²x²), so

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

**Partial fractions** (a ≠ 1):

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

```text
dI/da = [1/(1−a²)] [π/2 − a²·π/(2a)]
      = [1/(1−a²)] (π/2)(1 − a)
      = π / (2(1 + a))
```
Integrate: I(a) = (π/2) log(1 + a) + C. At a = 0, I = 0, so C = 0.

**Answer: ∫₀^∞ tan⁻¹(ax) / (x(1 + x²)) dx = (π/2) log(1 + a).**

## Check yourself

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

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

2. dI/da evaluates to
   A. π/(2(1+a))
   B. π/(2(1-a))
   C. π/(1+a)
   D. π a/(2(1+a))
   Answer: A. After partial fractions, (π/2)(1-a)/(1-a^2). (Key point: dI/da)

3. The integral equals
   A. (π/2) tan^-1 a
   B. (π/2) log(1+a)
   C. π log(1+a)
   D. (π/2) log a
   Answer: B. Integrating and I(0)=0 gives C = 0. (Key point: Result)

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