Question:
The value of \( \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx \) is
The value of \( \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx \) is
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Always check for odd functions when the integration limits are of the form \( [-a, a] \).
Odd function times even function is always an odd function. Here, \( x^3 \) is odd and the denominator is even.
Odd function times even function is always an odd function. Here, \( x^3 \) is odd and the denominator is even.
Updated On: Sep 10, 2026
- \( 0 \)
- \( \log 2 \)
- \( 2 \log 2 \)
- \( \frac{1}{2} \log 2 \)
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The Correct Option is A
Solution and Explanation
Concept:
• Property of Definite Integrals: \( \int_{-a}^{a} f(x) dx = 0 \) if \( f(x) \) is an odd function.
• A function is odd if \( f(-x) = -f(x) \).
• A function is even if \( f(-x) = f(x) \).
Step 1: Define the integrand function and test for symmetry
Let the integrand be \( f(x) \):
\[ f(x) = \frac{x^3}{x^2 + 2|x| + 1} \] To check if the function is odd or even, replace \( x \) with \( -x \):
\[ f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} \]
Step 2: Simplify the expression for \( f(-x) \)
Using the properties \( (-x)^3 = -x^3 \), \( (-x)^2 = x^2 \), and \( |-x| = |x| \):
\[ f(-x) = \frac{-x^3}{x^2 + 2|x| + 1} \] \[ f(-x) = - \left( \frac{x^3}{x^2 + 2|x| + 1} \right) \] \[ f(-x) = -f(x) \]
This proves that \( f(x) \) is an odd function.
Step 3: Apply the definite integral property
Since the function is odd and the limits of integration are symmetric about the origin (\( -1 \) to \( 1 \)):
\[ \int_{-1}^{1} f(x) dx = 0 \] Therefore:
\[ \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx = 0 \]
• Property of Definite Integrals: \( \int_{-a}^{a} f(x) dx = 0 \) if \( f(x) \) is an odd function.
• A function is odd if \( f(-x) = -f(x) \).
• A function is even if \( f(-x) = f(x) \).
Step 1: Define the integrand function and test for symmetry
Let the integrand be \( f(x) \):
\[ f(x) = \frac{x^3}{x^2 + 2|x| + 1} \] To check if the function is odd or even, replace \( x \) with \( -x \):
\[ f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} \]
Step 2: Simplify the expression for \( f(-x) \)
Using the properties \( (-x)^3 = -x^3 \), \( (-x)^2 = x^2 \), and \( |-x| = |x| \):
\[ f(-x) = \frac{-x^3}{x^2 + 2|x| + 1} \] \[ f(-x) = - \left( \frac{x^3}{x^2 + 2|x| + 1} \right) \] \[ f(-x) = -f(x) \]
This proves that \( f(x) \) is an odd function.
Step 3: Apply the definite integral property
Since the function is odd and the limits of integration are symmetric about the origin (\( -1 \) to \( 1 \)):
\[ \int_{-1}^{1} f(x) dx = 0 \] Therefore:
\[ \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx = 0 \]
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