Deadly integral $\int_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$. How to solve this question

$$\int\limits_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$$

. Please help me in solving this short way 
my approach is in the answer
Is it correct and can it be solved in a shorter way ?
 A: Multiply top and bottom by $(-(x-1)$, we have $\displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx$
Notice that the integrand can be written as a sum of a convergent geometric series with common ratio  $x^5$
$\begin{eqnarray} \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx &=& \displaystyle \int_0^1 \displaystyle \sum_{r=0}^\infty (1-x^3) x^{5r} \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \int_0^1 (x^{5r} - x^{3+5r} ) \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \left( \frac1{5r+1} - \frac1{5r+4}\right ) \\ \displaystyle &=& \sum_{r=0}^\infty \frac3{25r^2+25r+4} \qquad (\star) \\ \end{eqnarray}$
We consider the logarithmic differentiation of the Weiestrass Product:
$[\displaystyle \cos(\pi x) = \prod_{n=0}^\infty \left( 1 - \frac{4x^2}{(2n+1)^2} \right)$
to get
$\displaystyle \sum_{n=0}^\infty \frac1{(2n+1)^2 - (2x)^2} = \frac{\pi}{8x} \tan(\pi x ) \qquad (\star \star)$
Notice that  $ \frac3{25r^2+25r+4} = \frac{12}{25} \cdot \frac1{(2r+1)^2 - \left (2\cdot \frac3{10} \right)^2}$ Thus $x=\frac3{10}$ for $(\star \star)$
Hence, the integral in question equals to  $\dfrac{12}{25} \cdot \dfrac{\pi}{8\cdot 3/10} \tan\left(\pi \cdot \dfrac3{10}\right) = \dfrac\pi5 \tan\left(\dfrac3{10}\pi\right)$
Now, we just need to evaluate  $\tan\left(\frac3{10}\pi\right)$ .Apply the identity $\tan(x) = \sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}}$ for $x = \frac3{10}\pi$
This means we need to find what is the value of $\cos\left(\frac35\pi \right)$ Let $y$ denote this value. Because  $\cos\left(\frac{1}5\pi\right) = -\cos\left( 4\times \frac15\pi \right)$. Apply the double angle formula twice yields $\cos\left( \frac15\pi\right) = \frac{1+\sqrt5}4$. Apply the triple angle formula to get  $y = \frac{1-\sqrt5}4$
Substitution yields the answer of $\displaystyle \pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648 \ \square$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}
\,\dd x} =
\int_{0}^{1}{\pars{x^{2} + x + 1}\pars{x - 1} \over
             \pars{x^{4} + x^{3} + x^{2} + x + 1}\pars{x - 1}}\,\dd x =
\int_{0}^{1}{1 - x^{3} \over 1 - x^{5}}\,\dd x
\\[3mm] \stackrel{x^{5}\ \to x}{=} &
\int_{0}^{1}{1 - x^{3/5} \over 1 - x}\,{1 \over 5}\,x^{-4/5}\,\dd x =
{1 \over 5}\bracks{%
\int_{0}^{1}{1 - x^{-1/5} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-4/5} \over 1 - x}\,\dd x}
\end{align}
However,
$\ds{\Psi\pars{z} + \gamma = \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\ \Re\pars{z} > 0}$. $\Psi$ and $\gamma$ are the digamma function and the
Euler-Mascheroni constant, respectively. See, for example, Abramowitz $\&$ Stegun Table. Then,
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}
\,\dd x} =
{1 \over 5}\braces{\bracks{\Psi\pars{-\,{1 \over 5} + 1} + \gamma} -
\bracks{\Psi\pars{-\,{4 \over 5} + 1} + \gamma}}
\\[3mm] = &
{1 \over 5}\bracks{\Psi\pars{4 \over 5} - \Psi\pars{1 \over 5}} =
{1 \over 5}\,\pi\cot\pars{\pi\,{1 \over 5}}
\end{align}
In the last step we used the PolyGamma's Euler identity.
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}
\,\dd x} =
{1 \over 5}\,\pi\cot\pars{\pi \over 5} =
{1 \over 5}\,\pi\root{1 + {2 \over \root{5}}}
\\[3mm] = &
\color{#f00}{{1 \over 25}\,\pi\root{10\root{5} + 25}} \approx 0.8648
\end{align}
