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 use the property of definite integrals for even functions. A function $f(x)$ is even if \( f(-x) = f(x) \). For such functions, \( \int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx \).

Step 2: Detailed Explanation:
Let \( f(x) = \frac{\log_e(1 + |x|)}{1 + |x|} \).
Since \( |-x| = |x| \), it follows that \( f(-x) = f(x) \). Thus, $f(x)$ is an even function.
\[ I = \int_{-1}^1 \frac{\log_e(1 + |x|)}{1 + |x|} dx = 2 \int_0^1 \frac{\log_e(1 + x)}{1 + x} dx \]
For the integral on the right, use substitution:
Let \( u = \log_e(1 + x) \implies du = \frac{1}{1 + x} dx \).
Change the limits:
When \( x = 0 \), \( u = \log_e 1 = 0 \).
When \( x = 1 \), \( u = \log_e 2 \).
Substitute into the integral:
\[ I = 2 \int_0^{\log_e 2} u du = 2 \left[ \frac{u^2}{2} \right]_0^{\log_e 2} \]
\[ I = 2 \left( \frac{(\log_e 2)^2}{2} - 0 \right) = (\log_e 2)^2 \]

Step 3: Final Answer:
The value of the integral is \( (\log_e 2)^2 \).