Mathematics · Definite Integration

JEE Main 2026 — 6 April, Evening Shift — Question 39

The value of the integral $\int_{-1}^{1}\frac{x^{3}+|x|+1}{x^{2}+2|x|+1}dx$ is equal to :

Option A:
$3 \log _{e} 2$
Option B:
$2 \log _{e} 2$
Correct
Option C:
$5 \log _{e} 3$
Option D:
$3 \log _{e} 3$

Answer: B

Step-by-step solution

$I=\int_{-1}^{1} \frac{x^{3}+|x|+1}{x^{2}+2|x|+1} . d x$ $\mathrm{I}=\int_{-1}^{1} \frac{\mathrm{x}^{3}}{\mathrm{x}^{2}+2|\mathrm{x}|+1} \cdot \mathrm{dx}+\int_{-1}^{1} \frac{|\mathrm{x}|+1}{\mathrm{x}^{2}+2|\mathrm{x}|+1} \cdot \mathrm{dx}$ $\mathrm{I}=0+\int_{-1}^{1} \frac{|\mathrm{x}|+1}{|\mathrm{x}|^{2}+2|\mathrm{x}|+1} . \mathrm{dx}$ $2 \int_{0}^{1} \frac{1}{x+1} \cdot d x$ $\mathrm{I}=2[\ln |\mathrm{x}+1|]_{0}^{1}$ $\mathrm{I}=2 \ell \mathrm{n}(2)$

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Mathematics
Chapter: Definite Integration
Topic: Methods of solving definite integrals(kings rule,odd even)