Question

The value of \[ \int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \, dx is equal to: \]

• A. \( \log 2 \)
• B. \( 2 \log 2 \)
• C. \( \frac{1}{2} \log 2 \)
• D. \( 4 \log 2 \)

Hint:

For integrals with \( |x| \): \begin{itemize} \item Split at 0. \item Use symmetry substitutions like \( x \to -x \). \item Look for denominator squares. \end{itemize}

Answer 1

The Correct Option is B

Concept:
Break the integral at \( x = 0 \) due to absolute values: \[ |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \] Also: \[ x^2 + 2|x| + 1 = (|x| + 1)^2 \]
Step 1: Split the integral.
\[ I = \int_{-1}^{0} \frac{x^3 - x + 1}{(1 - x)^2} \, dx + \int_{0}^{1} \frac{x^3 + x + 1}{(x + 1)^2} \, dx \]
Step 2: Use substitution symmetry.
Let \( x \to -x \) in the first integral: \[ \int_{0}^{1} \frac{-x^3 + x + 1}{(1 + x)^2} \, dx \] Add both integrals: \[ I = \int_{0}^{1} \frac{2(x + 1)}{(x + 1)^2} \, dx \] \[ = \int_{0}^{1} \frac{2}{x + 1} \, dx \]
Step 3: Evaluate.
\[ I = 2 \int_{0}^{1} \frac{1}{x + 1} \, dx \] \[ = 2 [\ln(x + 1)]_{0}^{1} \] \[ = 2 \ln 2 \]