---
title: "Triple integral 1/(x+y+z+1)³"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:triple-integral-1-x-y"
source: https://www.bctengineer.com/q/semester_2/mathematics/triple-integral-1-x-y
site: BCT Engineering
---

# Triple integral 1/(x+y+z+1)³

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2071 Bhadra, 5 marks
- Video: [Triple Integral 1/(x+y+z)³ | Variable Limits Problem](https://www.youtube.com/watch?v=6dZoIWHSs7I) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/triple-integral-1-x-y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate: ∫₀¹ ∫₀^(1−x) ∫₀^(1−x−y) dx dy dz/(x+y+z+1)³.

## Short answer

Integrate in the order z, y, x. The z-integral gives (1/2)[1/(x+y+1)² − 1/4]; further integrations yield (1/2)ln2 − 5/16.

## Answer

**Reading.** The integrand is written with dx dy dz as the differentials, but the limits are for z innermost (0 to 1−x−y), then y (0 to 1−x), then x (0 to 1). I integrate z first, then y, then x.

I = ∫₀¹ ∫₀^{1−x} ∫₀^{1−x−y} dz dy dx / (x + y + z + 1)³

#### z-integration
```text
∫₀^{1−x−y} (x+y+z+1)⁻³ dz = [ −1/(2(x+y+z+1)²) ]₀^{1−x−y}
                          = (1/2) [ 1/(x+y+1)² − 1/4 ]
```

(At z = 1 − x − y, x + y + z + 1 = 2.)

#### y-integration
```text
∫₀^{1−x} (1/2)[ (x+y+1)⁻² − 1/4 ] dy
 = (1/2) [ −1/(x+y+1) − y/4 ]₀^{1−x}
 = (1/2) [ −1/2 − (1−x)/4 + 1/(x+1) ]
```

#### x-integration
```text
I = (1/2) ∫₀¹ [ 1/(1+x) − 1/2 − (1−x)/4 ] dx
  = (1/2) [ ln 2 − 1/2 − 1/8 ]
  = (1/2) ln 2 − 5/16
```

**Answer: I = (1/2) ln 2 − 5/16 ≈ 0.0341.**

## Check yourself

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

1. Integrate first in:
   A. x
   B. No particular order is needed
   C. y
   D. z
   Answer: D. Limits indicate z innermost. (Key point: Order)

2. After z:
   A. (1/2)[1/(x+y+1)² − 1/4]
   B. −1/(x+y+1)²
   C. 1/(x+y+1)²
   D. (1/2)/(x+y+1)
   Answer: A. Antiderivative −1/(2s²). (Key point: z result)

3. Value:
   A. 5/16
   B. (1/2)ln 2
   C. ln 2 − 5/16
   D. (1/2)ln 2 − 5/16
   Answer: D. Final integration. (Key point: Result)

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