Question

Evaluate: \[ \int_{-2}^{2} \frac{x^3 + |x| + 1}{x^2 + 4|x| + 4} \, dx. \]

Hint:

Use the symmetry of the function to simplify definite integrals, especially when dealing with even and odd functions.

Answer 1

Step 1: Analyze the denominator and absolute value. The denominator \( x^2 + 4|x| + 4 \) depends on the value of \( x \): - For \( x \geq 0 \), \( |x| = x \), so \[ x^2 + 4|x| + 4 = x^2 + 4x + 4 = (x + 2)^2. \] - For \( x <0 \), \( |x| = -x \), so \[ x^2 + 4|x| + 4 = x^2 - 4x + 4 = (x - 2)^2. \] Step 2: Split the integral into two regions. \[ \int_{-2}^{2} \frac{x^3 + |x| + 1}{x^2 + 4|x| + 4} \, dx = \int_{-2}^{0} \frac{x^3 - x + 1}{(x - 2)^2} \, dx + \int_{0}^{2} \frac{x^3 + x + 1}{(x + 2)^2} \, dx. \] Step 3: Analyze the symmetry of the function. Consider the numerator \( x^3 + |x| + 1 \): - The function \( x^3 \) is odd. - The term \( |x| \) is even. - The constant term 1 is even. Evaluate the integral properties using the symmetry of the function. Since the function inside the integral has a symmetric structure, the integral simplifies to: \[ \int_{-2}^{2} \frac{x^3 + |x| + 1}{x^2 + 4|x| + 4} \, dx = 0. \] Final Answer: \[ \int_{-2}^{2} \frac{x^3 + |x| + 1}{x^2 + 4|x| + 4} \, dx = 0. \]