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: Simplifying the Integrand:
Note that $x^4 + x^2 + 1 = (x^2+x+1)(x^2-x+1)$, so: \[ \frac{x(x^2+x+1)}{(x+1)(x^4+x^2+1)} = \frac{x(x^2+x+1)}{(x+1)(x^2+x+1)(x^2-x+1)} = \frac{x}{(x+1)(x^2-x+1)} \] Also note $(x+1)(x^2-x+1) = x^3+1$, so: \[ I = \int_0^2 \sqrt{\frac{x}{x^3+1}}\,dx = \int_0^2 \frac{\sqrt{x}}{\sqrt{x^3+1}}\,dx = \int_0^2 \frac{x^{1/2}}{\sqrt{x^3+1}}\,dx \] Rewrite as: \[ I = \int_0^2 \sqrt{\frac{x}{x^3+1}}\,dx \] Step 2: Substitution:
Let $t = x^{3/2}$, so $dt = \dfrac{3}{2}x^{1/2}\,dx$, i.e., $x^{1/2}\,dx = \dfrac{2}{3}\,dt$.
When $x=0$: $t=0$; when $x=2$: $t=2^{3/2}$.
Also $x^3 = t^2$, so: \[ I = \int_0^{2^{3/2}} \frac{1}{\sqrt{t^2+1}}\cdot\frac{2}{3}\,dt = \frac{2}{3}\int_0^{2^{3/2}} \frac{dt}{\sqrt{t^2+1}} \] Step 3: Evaluating the Standard Integral:
Recall $\displaystyle\int \dfrac{dt}{\sqrt{t^2+1}} = \ln\!\left(t + \sqrt{t^2+1}\right) + C$.
\[ I = \frac{2}{3}\Big[\ln\!\left(t+\sqrt{t^2+1}\right)\Big]_0^{2^{3/2}} \] \[ = \frac{2}{3}\left[\ln\!\left(2^{3/2} + \sqrt{2^3+1}\right) - \ln\!\left(0+\sqrt{0+1}\right)\right] \] \[ = \frac{2}{3}\left[\ln\!\left(2^{3/2}+\sqrt{9}\right) - \ln(1)\right] \] \[ = \frac{2}{3}\ln\!\left(2^{3/2}+3\right) \] Step 4: Final Answer:
\[ I = \frac{2}{3}\ln\!\left(2^{3/2}+3\right) \] The answer is Option (4).