Question

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

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

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The integral is defined over a symmetric interval \([-1, 1]\). We can split the integrand into odd and even parts. The integral of an odd function over a symmetric interval is zero, whereas the integral of an even function is twice the integral from 0 to 1.
Step 2: Key Formula or Approach:
Split the integral as: \[ I = \int_{-1}^{1} \frac{x^3}{x^2 + |x| + 1} dx + \int_{-1}^{1} \frac{|x| + 1}{x^2 + |x| + 1} dx \] The first part involves an odd function \( f(x) = \frac{x^3}{x^2 + |x| + 1} \), so its integral is 0. The second part involves an even function, so: \[ I = 2 \int_{0}^{1} \frac{x + 1}{x^2 + x + 1} dx \]
Step 3: Detailed Explanation:
To integrate \( \frac{x+1}{x^2+x+1} \), we write the numerator in terms of the derivative of the denominator \( (2x+1) \): \[ I = \int_{0}^{1} \frac{(2x + 1) + 1}{x^2 + x + 1} dx \] \[ I = \int_{0}^{1} \frac{2x + 1}{x^2 + x + 1} dx + \int_{0}^{1} \frac{1}{(x + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dx \] \[ I = [\ln(x^2 + x + 1)]_0^1 + \left[ \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \right]_0^1 \] \[ I = (\ln 3 - \ln 1) + \frac{2}{\sqrt{3}} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(\frac{1}{\sqrt{3}}) \right) \] \[ I = \ln 3 + \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \ln 3 + \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} = \ln 3 + \frac{\pi}{3\sqrt{3}} \]
Step 4: Final Answer:
The value of the integral is \( \ln 3 + \dfrac{\pi}{3\sqrt{3}} \).