Question

Sol. [Key Concept: Substitute $x^{3/2} = t$ to reduce the integral to $\frac{2}{3}\int\frac{dt}{\sqrt{t^{2} + 1}}$, giving $\frac{2}{3}ln(3 + 2\sqrt{2)}$.]

$$I = \int_{0}^{2}\mspace{2mu}\sqrt{\frac{x\left( x^{2} + x + 1 \right)}{(x + 1)\left( x^{2} - x + 1 \right)\left( x^{2} + x + 1 \right)}dx}$$

$$\begin{matrix} & \ = \int_{0}^{2}\mspace{2mu}\mspace{2mu}\sqrt{\frac{x}{x^{3} + 1}}dx \\ & \left( x^{3/2} = t \Rightarrow \frac{3}{2}x^{1/2}dx = dt \right) \\ & \ = \frac{2}{3}\int_{0}^{2^{3/2}}\mspace{2mu}\mspace{2mu}\frac{dt}{\sqrt{t^{2} + 1}} \\ & \ = \frac{2}{3}\left( ln\left( t + \sqrt{t^{2} + 1} \right) \right)_{0}^{2^{3/2}} \\ & \ = \frac{2}{3}ln\left( 2^{3/2} + 3 \right) \end{matrix}$$