Question

The value of $\int_{-1}^{1} \frac{\log_e(1 + |x|)}{1 + |x|} dx$ is equal to

• A. $\frac{1}{2} (\log_e 2)^2$
• B. $2 (\log_e 2)^2$
• C. $(\log_e 2)^2$
• D. $\log_e 2$
• E. 0

Hint:

Whenever an absolute value \( |x| \) is the only variable part of an integrand over a symmetric interval \([-a, a]\), the function is even. Simplify the bounds to \([0, a]\) immediately to get rid of the modulus sign.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are dealing with a definite integral over a symmetric interval \([-1, 1]\). The integrand involves an absolute value, making it an even function.
Step 2: Key Formula or Approach:
For an even function \(f(-x) = f(x)\), the integral property holds:
\(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\).
Once reduced to \([0,1]\), we can drop the absolute value and use substitution.
Step 3: Detailed Explanation:
Let \(f(x) = \frac{\log_e(1 + |x|)}{1 + |x|}\).
Clearly, \(f(-x) = f(x)\), so the function is even.
Apply the symmetry property:
\[ I = 2 \int_{0}^{1} \frac{\log_e(1 + x)}{1 + x} dx \] Use substitution. Let \(u = \log_e(1 + x)\).
Differentiate: \(du = \frac{1}{1 + x} dx\).
Change the limits of integration:
When \(x = 0, u = \log_e(1) = 0\).
When \(x = 1, u = \log_e(2)\).
The integral becomes:
\[ I = 2 \int_{0}^{\log_e 2} u \, du \] Evaluate the integral:
\[ I = 2 \left[ \frac{u^2}{2} \right]_{0}^{\log_e 2} \] \[ I = \left[ u^2 \right]_{0}^{\log_e 2} = (\log_e 2)^2 - 0^2 = (\log_e 2)^2 \] Step 4: Final Answer:
The value is \((\log_e 2)^2\).