---
title: "Integral of √x/(√x+√(a−x))"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-sqrt-x-over-sum-roots"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-sqrt-x-over-sum-roots
site: BCT Engineering
---

# Integral of √x/(√x+√(a−x))

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2071 Shrawan, 5 marks
- Video: [Integrate √x/(√x+√a-x) from 0 to a || Calculus || Engineering Mathematics](https://www.youtube.com/watch?v=3MwCD1qSVf4) — Mathematics Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-sqrt-x-over-sum-roots
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Evaluate ∫₀^a √x/(√x + √(a−x)) dx.

## Short answer

Apply ∫_0^a f(x)dx = ∫_0^a f(a-x)dx to write I = ∫_0^a √(a-x)/(√(a-x)+√x)dx. Adding the two forms yields 2I = ∫_0^a dx = a, so I = a/2.

## Answer

Let I = ∫₀^a √x/(√x + √(a − x)) dx. Using ∫₀^a f(x) dx = ∫₀^a f(a − x) dx:

I = ∫₀^a √(a − x)/(√(a − x) + √x) dx

Add the two forms:

2I = ∫₀^a (√x + √(a − x))/(√x + √(a − x)) dx = ∫₀^a dx = a

**Answer: I = a/2.**

## Check yourself

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

1. The property used is
   A. integral f(x) = 2 integral from 0 to a/2
   B. integral from 0 to a f(x) = integral from 0 to a f(a-x)
   C. integration by parts
   D. substitution x = a sin^2 θ
   Answer: B. This produces a second form of I. (Key point: Property)

2. Replacing x by a - x gives
   A. sqrt x/(sqrt x - sqrt(a-x))
   B. sqrt(a-x)/(sqrt(a-x) - sqrt x)
   C. sqrt(a-x)/(sqrt(a-x)+sqrt x)
   D. sqrt x/(sqrt(a-x)+sqrt x)
   Answer: C. Roots swap. (Key point: Second form)

3. The sum of the two forms has integrand
   A. x
   B. sqrt x
   C. a
   D. 1
   Answer: D. Numerator equals denominator. (Key point: Adding)

4. I equals
   A. a/2
   B. a
   C. a^2/2
   D. a/4
   Answer: A. 2I = a. (Key point: Result)

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