Question

The value of the integral \(\int_{0}^{2} \frac{\sqrt{x(x^2 + x + 1)}}{(\sqrt{x+1})(\sqrt{x^4 + x^2 + 1})} \, dx\) is equal to:

• A. \(\frac{1}{3} \log_e (3 - 2\sqrt{2})\)
• B. \(\frac{2}{3} \log_e (4 + \sqrt{2})\)
• C. \(\frac{2}{3} \log_e (3 + 2\sqrt{2})\)
• D. \(\frac{1}{3} \log_e (1 + 6\sqrt{2})\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This integral involves a complex radical expression. The key is to simplify the denominator using the factorization of \(x^4 + x^2 + 1\), which is \((x^2 + x + 1)(x^2 - x + 1)\).
Step 2: Key Formula or Approach:
1. Factorization: \(x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\).
2. Substitution: Simplify the integrand to a form that allows for a standard logarithmic integration.
Step 3: Detailed Explanation:
The integrand is: \[ I = \int_{0}^{2} \frac{\sqrt{x}\sqrt{x^2+x+1}}{\sqrt{x+1}\sqrt{x^2+x+1}\sqrt{x^2-x+1}} dx = \int_{0}^{2} \frac{\sqrt{x}}{\sqrt{x+1}\sqrt{x^2-x+1}} dx \] \[ I = \int_{0}^{2} \frac{\sqrt{x}}{\sqrt{x^3+1}} dx \] Let \(x^{3/2} = t \implies \frac{3}{2}x^{1/2} dx = dt \implies \sqrt{x} dx = \frac{2}{3} dt\). Limits: When \(x=0, t=0\). When \(x=2, t=2^{3/2} = 2\sqrt{2}\). \[ I = \frac{2}{3} \int_{0}^{2\sqrt{2}} \frac{dt}{\sqrt{t^2+1}} \] \[ I = \frac{2}{3} [\ln|t + \sqrt{t^2+1}|]_0^{2\sqrt{2}} \] \[ I = \frac{2}{3} [\ln(2\sqrt{2} + \sqrt{8+1}) - \ln(1)] = \frac{2}{3} \ln(3 + 2\sqrt{2}) \]
Step 4: Final Answer:
The value of the integral is \(\frac{2}{3} \log_e (3 + 2\sqrt{2})\).