Question

The value of the integral \[ \int_{-1}^{1}\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right)dx \] is equal to :

• A. \(3\log_e2\)
• B. \(2\log_e2\)
• C. \(5\log_e3\)
• D. \(3\log_e3\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We need to evaluate a definite integral over a symmetric interval \( [-1, 1] \). The integrand contains absolute value functions, which suggests that properties of even and odd functions might be useful.
Step 2: Key Formula or Approach
1. Property of definite integrals over symmetric intervals: \( \int_{-a}^{a} f(x) dx = \int_{-a}^{a} f_{even}(x) dx + \int_{-a}^{a} f_{odd}(x) dx \).
- If \(h(x)\) is an odd function (\(h(-x) = -h(x)\)), then \( \int_{-a}^{a} h(x) dx = 0 \).
- If \(h(x)\) is an even function (\(h(-x) = h(x)\)), then \( \int_{-a}^{a} h(x) dx = 2 \int_{0}^{a} h(x) dx \).
2. The denominator can be simplified: \( x^2 + 2|x| + 1 = |x|^2 + 2|x| + 1 = (|x| + 1)^2 \).
Step 3: Detailed Explanation
Let the integral be \(I\). The integrand is \( f(x) = \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \).
Using the simplified denominator, \( f(x) = \frac{x^3 + |x| + 1}{(|x| + 1)^2} \).
Let's split the integrand into parts: \[ I = \int_{-1}^{1} \frac{x^3}{(|x| + 1)^2} dx + \int_{-1}^{1} \frac{|x| + 1}{(|x| + 1)^2} dx \] Let's analyze the first integral. The integrand is \( h(x) = \frac{x^3}{(|x| + 1)^2} \). Check if it's odd or even: \[ h(-x) = \frac{(-x)^3}{(|-x| + 1)^2} = \frac{-x^3}{(|x| + 1)^2} = -h(x) \] Since \(h(x)\) is an odd function, its integral over the symmetric interval \( [-1, 1] \) is 0. \[ \int_{-1}^{1} \frac{x^3}{(|x| + 1)^2} dx = 0 \] Now, let's look at the second integral: \[ I = \int_{-1}^{1} \frac{|x| + 1}{(|x| + 1)^2} dx = \int_{-1}^{1} \frac{1}{|x| + 1} dx \] The integrand here is \( k(x) = \frac{1}{|x| + 1} \). Check if it's odd or even: \[ k(-x) = \frac{1}{|-x| + 1} = \frac{1}{|x| + 1} = k(x) \] Since \(k(x)\) is an even function, we can simplify the integral: \[ I = 2 \int_{0}^{1} \frac{1}{|x| + 1} dx \] For \( x \in [0, 1] \), \( |x| = x \). \[ I = 2 \int_{0}^{1} \frac{1}{x + 1} dx \] Now, we evaluate this standard integral: \[ I = 2 \left[ \ln|x + 1| \right]_0^1 \] \[ I = 2 (\ln(1 + 1) - \ln(0 + 1)) = 2 (\ln(2) - \ln(1)) \] Since \( \ln(1) = 0 \), we get: \[ I = 2 \ln(2) = 2 \log_e 2 \] Step 4: Final Answer
The value of the integral is \( 2 \log_e 2 \).