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...
 A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
I &\equiv \int_{0}^{3}{x\root{x + 1}\,dx \over x^{2} + x + 1}
=\int_{0}^{3}x\root{x + 1}\,
\pars{{1 \over x + 1/2 - \root{3}\ic/2} - {1 \over x + 1/2 + \root{3}\ic/2}}\,
{1 \over \root{3}\ic}\,dx
\\[3mm]&={2\root{3} \over 3}\Im\int_{0}^{3}
{x\ \overbrace{\root{x + 1}}^{\ds{\equiv\ t}} \over x + 1/2 - \root{3}\ic/2}\,\dd x
={2\root{3} \over 3}\Im\int_{1}^{2}
{\pars{t^{2} - 1}t \over \pars{t^{2} - 1} + 1/2 - \root{3}\ic/2}\,2t\,\dd t
\\[3mm]&={4\root{3} \over 3}\Im\int_{1}^{2}
{\pars{t^{2} - 1}t^{2} \over t^{2} + z}\,\dd t
\qquad\mbox{where}\qquad
z \equiv -\,\half - {\root{3} \over 2}\,\ic =\expo{4\pi\ic/3}
\qquad\qquad\qquad\qquad\qquad\pars{1}
\end{align}
$z$ is a root of $z^{2} + z + 1 = 0$.

\begin{align}
I &={4\root{3} \over 3}\Im\int_{0}^{2}
{\bracks{\pars{t^{2} + z} - \pars{z + 1}}\bracks{\pars{t^{2} + z} - z}
\over t^{2} + z}\,\dd t
\\[3mm]&={4\root{3} \over 3}\Im\int_{0}^{2}
{\pars{t^{2} + z}^{2} - \pars{2z + 1}\pars{t^{2} + z}
 +\ \overbrace{z\pars{z + 1}}^{\ds{=\ - 1}} \over t^{2} + z}\,\dd t
\\[3mm]&={4\root{3} \over 3}\Im\int_{0}^{2}
\bracks{\pars{t^{2} + z} - \pars{2z + 1} - {1 \over t^{2} + z}}\,\dd t
={4\root{3} \over 3}\pars{{\root{3} \over 2}
                          - \Im\int_{0}^{2}{\dd t \over t^{2} + z}}
\\[3mm]&=2 - {4\root{3} \over 3}\,
\Im\pars{\expo{-2\pi\ic/3}\int_{0}^{2\expo{2\pi\ic/3}}{\dd t \over t^{2} + 1}}
\end{align}

$$
\mbox{Evaluate the right hand side}:\quad
I=2 - {4\root{3} \over 3}\,
\Im\bracks{\expo{-2\pi\ic/3}\arctan\pars{2\expo{2\pi\ic/3}}}
$$
