Class CBSE Class 12 Mathematics Definite Integrals Q #1846
COMPETENCY BASED
EVALUATE
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
The value of $\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx$ is
(A) 0
(B) log 2
(C) 2 log 2
(D) $\frac{1}{2}$ log 2
Correct Answer: A

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Detailed Solution

Step 1: Analyze the integrand

Let $f(x) = \frac{x^3}{x^2 + 2|x| + 1}$. We observe the symmetry of the function over the interval $[-1, 1]$.

Step 2: Check for Odd or Even function

Recall that $f(x)$ is an odd function if $f(-x) = -f(x)$. Let us evaluate $f(-x)$:

$$f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} = \frac{-x^3}{x^2 + 2|x| + 1} = -f(x)$$

Since $f(-x) = -f(x)$, the function is an odd function.

Step 3: Apply the property of definite integrals

For an odd function $f(x)$, the property of definite integrals states that:

$$\int_{-a}^{a} f(x) dx = 0$$

Applying this to our integral where $a = 1$:

$$\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx = 0$$

Final Answer: 0

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Pedagogical Audit
Bloom's Analysis: This is an EVALUATE question because the student must analyze the properties of the integrand (symmetry/parity) to determine the result without performing complex integration.
Knowledge Dimension: PROCEDURAL
Justification: The student must apply the specific algorithm of checking function parity to simplify the definite integral.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the conceptual understanding of definite integral properties (King's property/Parity) rather than rote calculation.