Question

\(\int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + |x| + 1} \, dx\) is equal to

• A. \(\ln 3 - \frac{\pi}{3\sqrt{3}}\)
• B. \(\frac{\pi}{3\sqrt{3}}\)
• C. \(\ln 3 + \frac{\pi}{3\sqrt{3}}\)
• D. \(-\frac{\pi}{3\sqrt{3}}\)

Hint:

Splitting an integral into odd and even components is the most powerful technique for evaluating definite integrals with symmetric limits (\(-a\) to \(a\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The integral is over a symmetric interval \([-1, 1]\). We split the integrand into odd and even parts.

Step 2: Key Formula or Approach:
1. \(\int_{-a}^a f(x) dx = 0\) if \(f(x)\) is odd. 2. \(\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx\) if \(f(x)\) is even.

Step 3: Detailed Explanation:
Split the integral: \[ I = \int_{-1}^1 \frac{x^3}{x^2 + |x| + 1} dx + \int_{-1}^1 \frac{|x| + 1}{x^2 + |x| + 1} dx \] The first part is an odd function, so it becomes \(0\). The second part is an even function: \[ I = 2 \int_0^1 \frac{x + 1}{x^2 + x + 1} dx \] Adjust the numerator to include the derivative of the denominator (\(2x+1\)): \[ I = \int_0^1 \frac{2x + 1 + 1}{x^2 + x + 1} dx = \int_0^1 \frac{2x+1}{x^2+x+1} dx + \int_0^1 \frac{1}{(x+1/2)^2 + (\sqrt{3}/2)^2} dx \] \[ I = [\ln(x^2+x+1)]_0^1 + \left[ \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{x+1/2}{\sqrt{3}/2} \right) \right]_0^1 \] \[ I = (\ln 3 - \ln 1) + \frac{2}{\sqrt{3}} \left[ \tan^{-1}(\sqrt{3}) - \tan^{-1}(1/\sqrt{3}) \right] \] \[ I = \ln 3 + \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \ln 3 + \frac{\pi}{3\sqrt{3}} \]

Step 4: Final Answer:
The value of the integral is \(\ln 3 + \frac{\pi}{3\sqrt{3}}\).