Question

The value of \[ \int_0^{2} \sqrt{\frac{x(x^2+x+1)}{(x+1)(x^4+x^2+1)}} \, dx \] is 

• A. \(\dfrac{1}{3} \ln(2^{1/2} + 3)\)
• B. \(\ln(2^{1/2} + 3)\)
• C. \(\dfrac{2}{3} \ln(2^{3/2} + 3)\)
• D. \(\dfrac{2}{3} \ln(2^{1/2} + 3)\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This integral requires simplifying the expression under the square root using the identity \(x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\). After cancellation, we use a substitution to reduce the integral to a standard logarithmic form.

Step 2: Key Formula or Approach:
1. Simplify: \(\frac{x^2+x+1}{x^4+x^2+1} = \frac{1}{x^2-x+1}\). 2. The integral becomes: \(\int_0^2 \sqrt{\frac{x}{(x+1)(x^2-x+1)}} dx = \int_0^2 \sqrt{\frac{x}{x^3+1}} dx\).

Step 3: Detailed Explanation:
1. Rewrite: \(I = \int_0^2 \frac{x^{1/2}}{\sqrt{x^3+1}} dx\). 2. Substitute \(t = x^{3/2} \implies dt = \frac{3}{2} x^{1/2} dx\). 3. Limits: \(x=0 \to t=0\); \(x=2 \to t=2^{3/2}\). 4. \(I = \frac{2}{3} \int_0^{2^{3/2}} \frac{1}{\sqrt{t^2+1}} dt\). 5. Using \(\int \frac{dt}{\sqrt{t^2+1}} = \ln|t + \sqrt{t^2+1}|\): \[ I = \frac{2}{3} \left[ \ln(t + \sqrt{t^2+1}) \right]_0^{2^{3/2}} \] \[ I = \frac{2}{3} \left( \ln(2^{3/2} + \sqrt{8+1}) - \ln(1) \right) = \frac{2}{3} \ln(2^{3/2} + 3) \]

Step 4: Final Answer:
The value of the integral is \(\frac{2}{3} \ln(2^{3/2} + 3)\).