<s>[INST] How to evaluate the integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$ Evaluating this integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$
I've tried:
Set : $\sqrt{x+1}=t\mapsto dx=2t\,dt \Longrightarrow I=2\int_1^2\frac{t^4-t^2}{t^4-t^2+1}dt=2\int_1^2(1-\frac{1}{t^4-t^2+1})dt$
And then, I set $(t^2-\frac{1}{2})=\frac{\sqrt{3}}{2}\tan v$ ... But I can't continue to solve this integral...
 [/INST] HINT:
As $\displaystyle t^4-t^2+1=(t^2+1)^2-3t^2=(t^2+\sqrt3t+1)(t^2-\sqrt3t+1)$
$$\frac1{t^4-t^2+1}=\frac1{2\sqrt3t}\left(\frac1{t^2-\sqrt3t+1}-\frac1{t^2+\sqrt3t+1}\right)$$
Using Trigonometric substitution ,
$$t^2-\sqrt3t+1=\frac{4t^2-4\sqrt3t+4}4=\frac{(2t-\sqrt3)^2+1}4$$
So set, $2t-\sqrt3=\tan\phi$ etc.
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