JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

The value of the integral

\int_{-1}^{1} \left( \frac{x^{3} + |x| + 1}{x^{2} + 2|x| + 1} \right) dx

is equal to

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3\log_{e}2

2\log_{e}2

5\log_{e}3

3\log_{e}3

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

3\log_{e}2

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 $x \ge 0$ , $\frac{|x|+1}{(|x|+1)^2} = \frac{1}{x+1}$ , and multiplying by $2$ from symmetry gives $2\log_e 2$ , not $3\log_e 2$ .

B ·

2\log_{e}2

None. 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 $2\int_{0}^{1} \frac{1}{x+1}\,dx = 2\log_{e}2$ .

C ·

5\log_{e}3

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 ·

3\log_{e}3

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 $I = \int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$ .

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: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int_{-1}^1 \frac{|x|+1}{(|x|+1)^2} dx$ Since $f(x) = \frac{x^3}{x^2+2|x|+1}$ is odd ( $f(-x) = -f(x)$ ), $\int_{-1}^1 f(x) dx = 0$ . For the second term, since the integrand is even: $I = 0 + 2 \int_0^1 \frac{1}{x+1} dx = 2 \left[ \ln(x+1) \right]_0^1 = 2 (\ln 2 - \ln 1) = 2 \ln 2$

✓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 $2\ln 2 \approx 1.386$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $2 \int_{0}^{1} \frac{1}{x+1}\,dx$ , which evaluates to $2\log_{e}2$ .

Why each option works or fails

A: $3\log_{e}2$ - 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 $x \ge 0$ , $\frac{|x|+1}{(|x|+1)^2} = \frac{1}{x+1}$ , and multiplying by $2$ from symmetry gives $2\log_e 2$ , not $3\log_e 2$ .
B · correct: $2\log_{e}2$ - None. 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 $2\int_{0}^{1} \frac{1}{x+1}\,dx = 2\log_{e}2$ .
C: $5\log_{e}3$ - 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: $3\log_{e}3$ - 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 $I = \int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$ .
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: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int_{-1}^1 \frac{|x|+1}{(|x|+1)^2} dx$ Since $f(x) = \frac{x^3}{x^2+2|x|+1}$ is odd ( $f(-x) = -f(x)$ ), $\int_{-1}^1 f(x) dx = 0$ . For the second term, since the integrand is even: $I = 0 + 2 \int_0^1 \frac{1}{x+1} dx = 2 \left[ \ln(x+1) \right]_0^1 = 2 (\ln 2 - \ln 1) = 2 \ln 2$
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 $[0, 1]$ , which is close to $2\ln 2 \approx 1.386$ .

Integral Calculus: JEE Main Mathematics Question with Solution

Splitting the integral using odd-even symmetry simplifies it to 2∫011x+1 dx2 \int_{0}^{1} \frac{1}{x+1}\,dx2∫01​x+11​dx, which evaluates to 2log⁡e22\log_{e}22loge​2.

01given

02goal

03approach

04execute

✓verify

Students also ask

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