JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite Integral of a Rational Function Involving Absolute Values and Symmetry

Q: Why does x^2 + 2|x| + 1 equal (|x| + 1)^2?

Because x^2 = |x|^2 for all real x, so x^2 + 2|x| + 1 = |x|^2 + 2|x| + 1 = (|x| + 1)^2.

The value of the integral integral from -1 to 1 of ((x^3 + |x| + 1)/(x^2 + 2|x| + 1)) dx is equal to

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A3log_e2B2log_e2C5log_e3D3log_e3

Step-by-step solutionView

Correct answer

Splitting the integral using odd-even symmetry simplifies it to $2 \int_{0}^{1} \frac{1}{x + 1} d x$ , which evaluates to $2 \log_{e} 2$ .

Option analysis

Why each option works or fails

A · 3log_e2

The student miscalculated the algebraic simplification of $\frac{x + 1}{(x + 1)^{2}}$ or multiplied by an incorrect constant factor during symmetry reduction. Observe that for $x \geq 0$ , $\frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}} = \frac{1}{x + 1}$ , and multiplying by $2$ from symmetry gives $2 \log_{e} 2$ , not $3 \log_{e} 2$ .

B · 2log_e2

This is the correct option. Split the integrand into odd and even components: the odd part $\frac{x^{3}}{(∣ x ∣ + 1)^{2}}$ integrates to $0$ , and the even part integrates to $2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 \log_{e} 2$ .

C · 5log_e3

The student incorrectly evaluated the limits of integration or substituted an erroneous base/argument into the logarithm. Evaluate the integral with the correct upper limit $x = 1$ , which gives $\log_{e} (1 + 1) - \log_{e} (0 + 1) = \log_{e} 2$ .

D · 3log_e3

The student mistakenly substituted $x = 2$ into the antiderivative or miscomputed $\log_{e} (x + 1)$ at the boundaries. The integral bounds are from $0$ to $1$ , yielding $\log_{e} (1 + 1) - \log_{e} (1) = \log_{e} 2$ .

Reviewed route

Solution

StepWorking

01Given

The integral to evaluate is $I = \int_{- 1}^{1} \frac{x^{3} + ∣ x ∣ + 1}{x^{2} + 2 ∣ x ∣ + 1} d x$ .

02Goal

Compute the exact value of $I$ .

03Approach

Split the numerator into odd and even components: $\frac{x^{3}}{x^{2} + 2 ∣ x ∣ + 1}$ is an odd function because the numerator is odd and the denominator is even, so its integral over $[- 1, 1]$ is zero. The remaining part $\frac{∣ x ∣ + 1}{x^{2} + 2 ∣ x ∣ + 1}$ is an even function, which can be simplified as $\frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}} = \frac{1}{∣ x ∣ + 1}$ and integrated over $[0, 1]$ multiplied by 2.

04Execute

Split the integral: $I = \int_{- 1}^{1} \frac{x^{3}}{x^{2} + 2 ∣ x ∣ + 1} d x + \int_{- 1}^{1} \frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}} d x$ Since $f (x) = \frac{x^{3}}{x^{2} + 2 ∣ x ∣ + 1}$ is odd ( $f (- x) = - f (x)$ ), $\int_{- 1}^{1} f (x) d x = 0$ . For the second term, since the integrand is even: $I = 0 + 2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 {[\ln (x + 1)]}_{0}^{1} = 2 (\ln 2 - \ln 1) = 2 \ln 2$

✓Verify

Check positive integrand behavior on $[0, 1]$ : at $x = 0$ , $f (0) = 1$ ; at $x = 1$ , $f (1) = 3 / 4$ . The average value is around $0.8$ , over a base of length $2$ , total area $\approx 1.6$ , which is close to $2 \ln 2 \approx 1.386$ .

Hints that build this answer step by step

How can the integral $\int_{- 1}^{1} \frac{x^{3} + ∣ x ∣ + 1}{x^{2} + 2 ∣ x ∣ + 1} d x$ be simplified using symmetry on $[- 1, 1]$ ?
Separate the odd part $\frac{x^{3}}{(∣ x ∣ + 1)^{2}}$ and even part $\frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}}$ .
After eliminating the odd part, what is the simplified form of the remaining integral?
$2 \int_{0}^{1} \frac{1}{x + 1} d x$
What is the final evaluated value of $2 \int_{0}^{1} \frac{1}{x + 1} d x$ ?
$2 \log_{e} 2$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does

x^{2} + 2 ∣ x ∣ + 1

equal

(∣ x ∣ + 1)^{2}

Because $x^{2} = ∣ x ∣^{2}$ for all real $x$ , so $x^{2} + 2 ∣ x ∣ + 1 = ∣ x ∣^{2} + 2 ∣ x ∣ + 1 = (∣ x ∣ + 1)^{2}$ .

Answer

Splitting the integral using odd-even symmetry simplifies it to $2 \int_{0}^{1} \frac{1}{x + 1} d x$ , which evaluates to $2 \log_{e} 2$ .

Why each option works or fails

A: 3log_e2 - The student miscalculated the algebraic simplification of $\frac{x + 1}{(x + 1)^{2}}$ or multiplied by an incorrect constant factor during symmetry reduction. Observe that for $x \geq 0$ , $\frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}} = \frac{1}{x + 1}$ , and multiplying by $2$ from symmetry gives $2 \log_{e} 2$ , not $3 \log_{e} 2$ .
B · correct: 2log_e2 - This is the correct option. Split the integrand into odd and even components: the odd part $\frac{x^{3}}{(∣ x ∣ + 1)^{2}}$ integrates to $0$ , and the even part integrates to $2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 \log_{e} 2$ .
C: 5log_e3 - The student incorrectly evaluated the limits of integration or substituted an erroneous base/argument into the logarithm. Evaluate the integral with the correct upper limit $x = 1$ , which gives $\log_{e} (1 + 1) - \log_{e} (0 + 1) = \log_{e} 2$ .
D: 3log_e3 - The student mistakenly substituted $x = 2$ into the antiderivative or miscomputed $\log_{e} (x + 1)$ at the boundaries. The integral bounds are from $0$ to $1$ , yielding $\log_{e} (1 + 1) - \log_{e} (1) = \log_{e} 2$ .

Step-by-step solution

Given: The integral to evaluate is $I = \int_{- 1}^{1} \frac{x^{3} + ∣ x ∣ + 1}{x^{2} + 2 ∣ x ∣ + 1} d x$ .
Goal: Compute the exact value of $I$ .
Approach: Split the numerator into odd and even components: $\frac{x^{3}}{x^{2} + 2 ∣ x ∣ + 1}$ is an odd function because the numerator is odd and the denominator is even, so its integral over $[- 1, 1]$ is zero. The remaining part $\frac{∣ x ∣ + 1}{x^{2} + 2 ∣ x ∣ + 1}$ is an even function, which can be simplified as $\frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}} = \frac{1}{∣ x ∣ + 1}$ and integrated over $[0, 1]$ multiplied by 2.
Execute: Split the integral: $I = \int_{- 1}^{1} \frac{x^{3}}{x^{2} + 2 ∣ x ∣ + 1} d x + \int_{- 1}^{1} \frac{∣ x ∣ + 1}{(∣ x ∣ + 1)^{2}} d x$ Since $f (x) = \frac{x^{3}}{x^{2} + 2 ∣ x ∣ + 1}$ is odd ( $f (- x) = - f (x)$ ), $\int_{- 1}^{1} f (x) d x = 0$ . For the second term, since the integrand is even: $I = 0 + 2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 {[\ln (x + 1)]}_{0}^{1} = 2 (\ln 2 - \ln 1) = 2 \ln 2$
Verify: Check positive integrand behavior on $[0, 1]$ : at $x = 0$ , $f (0) = 1$ ; at $x = 1$ , $f (1) = 3 / 4$ . The average value is around $0.8$ , over a base of length $2$ , total area $\approx 1.6$ , which is close to $2 \ln 2 \approx 1.386$ .

Definite Integral of a Rational Function Involving Absolute Values and Symmetry

What feels right?

Splitting the integral using odd-even symmetry simplifies it to $2 \int_{0}^{1} \frac{1}{x + 1} d x$ , which evaluates to $2 \log_{e} 2$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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