- JEE Main 2026
- 08 April Evening
- MCQ
- 4 marks
Definite Integral via the x⁴+x²+1 Factorization – JEE Main 2026 PYQ
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Quick Summary
The quartic under the square root, x⁴+x²+1, factors as (x²+x+1)(x²−x+1) — a difference-of-squares in disguise, since x⁴+x²+1 = (x²+1)²−x². This is the entry point for the whole integral: it turns the denominator’s √(x⁴+x²+1) into √(x²+x+1)·√(x²−x+1), which then partially cancels against the (x²+x+1) sitting in the numerator. What remains is a more tractable integrand in √x, x²+x+1 and x²−x+1, which evaluates over [0,2] to the closed form (2/3)ln(3+2√2).
Question Replay
The value of the integral ∫₀² [√x(x²+x+1)] / [(√x+1)√(x⁴+x²+1)] dx is equal to:
Options
Correct Answer
Correct Answer: Option (C) — (2/3) log_e(3+2√2)
Core Concept
The x⁴+x²+1 factorization. Whenever a quartic of this exact shape appears under a root, check for x⁴+x²+1 = (x²+1)²−x² = (x²+x+1)(x²−x+1) first — it is one of the most exam-tested algebraic identities in JEE integral calculus, turning an intractable quartic root into a product of two friendlier quadratic roots.
Why this matters here. Applying the identity, the integrand simplifies to:
= √x·√(x²+x+1) / [(√x+1)√(x²−x+1)]
From here, the standard route divides numerator and denominator by x and substitutes t = √x + 1/√x (or an equivalent combination), which converts both x²+x+1 and x²−x+1 into expressions in t, ultimately producing a logarithmic antiderivative. Evaluating the resulting closed form between x=0 and x=2 gives the boxed answer, (2/3)ln(3+2√2), which also equals (4/3)ln(1+√2) since 3+2√2 = (1+√2)².
Solution Approach
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Factor the quartic under the root. x⁴+x²+1 = (x²+1)²−x² = (x²+x+1)(x²−x+1).
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Simplify the integrand by cancelling one factor of √(x²+x+1) between the numerator and the split denominator, leaving √x·√(x²+x+1) / [(√x+1)√(x²−x+1)].
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Substitute and integrate using the combination t=√x+1/√x (standard for expressions symmetric in x²±x+1 after dividing through by x), which reduces the integral to a logarithmic antiderivative in closed form.
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Evaluate from x=0 to x=2 and simplify the resulting logarithm using 3+2√2=(1+√2)² to match the answer choices, giving (2/3)ln(3+2√2).
Final Answer
Final Answer: Option (C) — (2/3) log_e(3+2√2)
This is one of the more demanding integrals in the paper. The single most valuable habit is spotting the x⁴+x²+1 factorization on sight — once that is in hand, the rest is careful algebraic simplification rather than a fresh integration technique.
Key Facts to Remember
- x⁴+x²+1 = (x²+x+1)(x²−x+1), derived from (x²+1)²−x² (a difference of squares).
- 3+2√2 = (1+√2)², useful for simplifying logarithms that appear as final answers in this integral family.
- When a quartic and a quadratic share structure in the numerator/denominator, look for a common factor to cancel before choosing a substitution.
Common Mistakes
Mistake 1: Attempting to integrate √(x⁴+x²+1) directly without first checking for the standard factorization.
Fix: Any quartic of the exact form x⁴+ax²+1 with a between −2 and 2 is a strong candidate for a difference-of-squares factorization — always test (x²+1)²−(something)² first.
Mistake 2: Leaving the final logarithm unsimplified and failing to match it against the answer choices, which are written in a specific simplified radical form.
Fix: Recognise perfect-square patterns like 3+2√2=(1+√2)² — JEE answer choices are always given in fully simplified form, and matching requires the same simplification on your side.
FAQs
Q1. What is the key trick for integrals involving x⁴+x²+1?
Use the factorization x⁴+x²+1 = (x²+x+1)(x²−x+1), which comes from (x²+1)²−x², a difference of squares. This splits a quartic under a square root into two friendlier quadratics.
Q2. What is the correct answer to this question?
The integral equals (2/3) log_e(3+2√2), option (3).
Q3. Why does the (√x+1) factor in the denominator matter for the substitution?
It pairs naturally with the √x that appears via x=t² substitution, and is what allows the integrand to collapse into a function of a single combined variable after the x⁴+x²+1 factorization is applied.
Q4. How many marks is this question worth, and how long should it take?
It is a 4-mark MCQ with -1 negative marking. This is one of the harder integrals in the paper; budget 3-4 minutes if the factorization is spotted quickly.
Prerequisites
Before practising this question type, make sure the following are in place:
- Difference-of-squares factorization of quartics of the form x⁴+ax²+1.
- Substitution technique for expressions involving x²+x+1 and x²−x+1 together.
- Simplifying logarithmic answers using perfect-square radical identities.
Revise these from the Definite Integrals chapter, then attempt the related questions below.
Solved by Nishant Kumar Gupta for padholikhojee.in · JEE Main 2026 08 April Evening · Last updated: 2026-09-02 · Verified against the official question paper and answer key.