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: Factorize the expression inside the radical We use the identity \[ x^4+x^2+1=(x^2+x+1)(x^2-x+1) \] because \[ (x^2+x+1)(x^2-x+1)=x^4+x^2+1 \] Therefore, \[ \sqrt{x^4+x^2+1} = \sqrt{x^2+x+1}\sqrt{x^2-x+1} \] Substituting into the integral, \[ I=\int_0^2 \frac{\sqrt{x}\sqrt{x^2+x+1}} {\sqrt{x+1}\sqrt{x^2+x+1}\sqrt{x^2-x+1}}\,dx \] Canceling \(\sqrt{x^2+x+1}\), \[ I=\int_0^2 \frac{\sqrt{x}} {\sqrt{x+1}\sqrt{x^2-x+1}}\,dx \] Step 2: Simplify the denominator Observe that \[ (x+1)(x^2-x+1)=x^3+1 \] Hence, \[ \sqrt{x+1}\sqrt{x^2-x+1}=\sqrt{x^3+1} \] So the integral becomes \[ I=\int_0^2 \frac{\sqrt{x}}{\sqrt{x^3+1}}\,dx \] Step 3: Use substitution Let \[ t=x^{3/2} \] Then \[ dt=\frac{3}{2}x^{1/2}\,dx \] which gives \[ \sqrt{x}\,dx=\frac{2}{3}\,dt \] Also, \[ x^3=t^2 \] So \[ \sqrt{x^3+1}=\sqrt{t^2+1} \] Changing the limits: When \(x=0\), \[ t=0 \] When \(x=2\), \[ t=2^{3/2}=2\sqrt{2} \] Thus, \[ I=\frac{2}{3}\int_0^{2\sqrt{2}}\frac{dt}{\sqrt{t^2+1}} \] Step 4: Apply standard integral formula Using \[ \int \frac{dt}{\sqrt{t^2+1}} = \log_e\left|t+\sqrt{t^2+1}\right|+C \] we get \[ I= \frac{2}{3} \left[ \log_e\left(t+\sqrt{t^2+1}\right) \right]_0^{2\sqrt{2}} \] \[ I= \frac{2}{3} \left( \log_e\left(2\sqrt{2}+\sqrt{8+1}\right) - \log_e(1) \right) \] \[ I= \frac{2}{3} \log_e(2\sqrt{2}+3) \] Hence, \[ \boxed{ I=\frac{2}{3}\log_e(3+2\sqrt{2}) } \] Final Answer: \[ \boxed{ \frac{2}{3}\log_e(3+2\sqrt{2}) } \]