Question 61 markMCQSection A

If ∫040dx2x+1=log⁡k\int_{0}^{40} \frac{dx}{2x + 1} = \log k, then the value of k is :

Calculus (Integration)
Definite Integration

Options

(A)33
(B)99
(C)92\frac{9}{2}
(D)32\frac{3}{2}

Official answer

The correct option is B) 9.


Working:


  • ∫dx2x+1=12log⁡(2x+1)\int \frac{dx}{2x+1} = \frac{1}{2}\log(2x+1).
  • Evaluate from 0 to 40: 12[log⁡(2⋅40+1)−log⁡(2⋅0+1)]=12[log⁡81−log⁡1]\frac{1}{2}[\log(2 \cdot 40+1) - \log(2 \cdot 0+1)] = \frac{1}{2}[\log 81 - \log 1].
  • =12log⁡81=log⁡811/2=log⁡9= \frac{1}{2} \log 81 = \log 81^{1/2} = \log 9.
  • Comparing with log⁡k\log k gives k=9k = 9.
Answer keywords:definite integralintegral of 1/(2x+1)(1/2) log(2x+1)log 81log 9k = 9square root inside log

Marking scheme

  1. 1.1 mark: correct antiderivative 12log⁡(2x+1)\frac{1}{2}\log(2x+1), evaluation to 12log⁡81=log⁡9\frac{1}{2}\log 81 = \log 9, and k=9k = 9 (option B).

Hint

Integrate to 12log⁡(2x+1)\frac{1}{2}\log(2x+1); then 12log⁡81=log⁡811/2=log⁡9\frac{1}{2}\log 81 = \log 81^{1/2} = \log 9.

Quick oral answer

The integral is 12log⁡(2x+1)\frac{1}{2}\log(2x+1); from 0 to 40 that is 12log⁡81=log⁡9\frac{1}{2}\log 81 = \log 9, so k equals 9.

Common mistakes

  1. 1.Omitting the factor 12\frac{1}{2} from the antiderivative, giving log⁡81\log 81 and k=81k = 81.
  2. 2.Converting 12log⁡81\frac{1}{2}\log 81 wrongly to log⁡(81/2)\log(81/2) instead of log⁡811/2\log 81^{1/2}.
  3. 3.Miscalculating the upper limit: 2(40)+1=812(40)+1 = 81, not 80 or 41.

Analysis and explanation

This tests evaluation of a definite integral of the form ∫dxax+b\int \frac{dx}{ax+b} and log manipulation.


Concept:


  • ∫dx2x+1=12log⁡(2x+1)\int \frac{dx}{2x+1} = \frac{1}{2}\log(2x+1) because of the inner-derivative factor 2.
  • The coefficient 1/2 becomes a power inside the log: 12log⁡81=log⁡9\frac{1}{2}\log 81 = \log 9.

Why B is correct:


  • 12[log⁡81−log⁡1]=12log⁡81=log⁡9\frac{1}{2}[\log 81 - \log 1] = \frac{1}{2}\log 81 = \log 9, so k=9k = 9.

Why the distractors are wrong:


  • A (3): from 9=3\sqrt{9} = 3, mistaking 81=9\sqrt{81} = 9 for a further root.
  • C (92\frac{9}{2}): from forgetting to fold the 1/2 into the log and instead multiplying 9 by 1/2 wrongly, or halving 9.
  • D (32\frac{3}{2}): a compounded version of the same 1/2 and root errors.

Interesting facts

The rule ∫dxax+b=1alog⁡∣ax+b∣\int \frac{dx}{ax+b} = \frac{1}{a}\log|ax+b| is one of the most frequently used standard integrals in applied calculus.

Since 81=3481 = 3^4, 81=9=32\sqrt{81} = 9 = 3^2 — the powers of 3 make this integral evaluate to a clean logarithm.

Question details

Q6 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Calculus (Integration)
Topic
Definite Integration
Bloom's level
apply
Syllabus topic
Unit 3: Calculus — Definite Integrals and their evaluation
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Where does the 12\frac{1}{2} factor come from?

Because the derivative of the inner expression 2x+1 is 2, integrating 12x+1\frac{1}{2x+1} introduces a 1/2, giving 12log⁡(2x+1)\frac{1}{2}\log(2x+1).

How does 12log⁡81\frac{1}{2}\log 81 become log⁡9\log 9?

A coefficient in front of a log becomes a power inside it: 12log⁡81=log⁡811/2=log⁡81=log⁡9\frac{1}{2}\log 81 = \log 81^{1/2} = \log \sqrt{81} = \log 9.