Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$ Insane integral $$\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln x} \mathop{dx}$$
I know $x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}$ but does it help?  I think $u=\ln{x}$ might be necessary some point.
 A: Write this as:
$$\int_{0}^1\frac{(x^2-1)(x-1)\,dx}{(x^5-1)\ln x}$$
Note that $$\int_0^1 x^y\,dy = (x-1)/\ln x$$
Replacing this and converting the problem into a double integral gives
$$\int_{0}^1\int_0^1\frac{(x^2-1)}{(x^5-1)}x^y\,dx\, dy$$
Writing $\frac{1}{1-x^5} = \sum_{k=0}^\infty x^{5k}$
This becomes
$$\int_{0}^1\int_0^1(1-x^2)x^y\sum_{k=0}^\infty x^{5k}\, dx\, dy$$
Interchanging the summation and integration
$$ = \sum_{k=0}^\infty \int_{0}^1\int_0^1(1-x^2)x^yx^{5k}\,dx\,dy$$
Integrating with respect to $x$,
$$ = \sum_{k=0}^\infty \int_{0}^1 \left(\frac{1}{5k+y+1} - \frac{1}{5k+y+3}\right)\,dy$$
Integrating this, we get
$$I = \sum_{k=0}^\infty \ln \frac{(5k+2)(5k+3)}{(5k+1)(5k+4)}$$
Using the identity provided by @Nanayajitzuki in the comments:
$$\ln \prod_{k=0}^\infty \frac{(5k+2)(5k+3)}{(5k+1)(5k+4)} = \ln\left(\frac{\Gamma(1/5)\Gamma(4/5)}{\Gamma(2/5)\Gamma(3/5)}\right)$$
Using Euler's reflection formula, $\Gamma(z)\Gamma(1-z) = \pi /\sin (\pi z)$
$$ = \ln \frac{\sin (2\pi/5)}{\sin (\pi/5)}$$
$$ = \ln (2\cos (\pi/5))$$
$$ = \ln (\phi)$$
A: (edit for a little glitch in writing)
so, basically you want to solve
$$
\int_{0}^{1} \frac{(x^2-1)(x-1)}{(x^5-1)\ln x} \mathrm{d}x
$$
let
$$
I(s) = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{(x^5-1)\ln x} \mathrm{d}x
$$
take derivative of which
$$
\begin{aligned}
I'(s) & = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{x^5-1} \mathrm{d}x = \int_{0}^{1} \frac{x^{s+3}-x^{s+2}-x^{s+1}+x^{s}}{x^5-1} \mathrm{d}x\\
& = \frac1{5} \int_{0}^{1} \frac{x^{\tfrac{s-1}{5}}-x^{\tfrac{s-2}{5}}-x^{\tfrac{s-3}{5}}+x^{\tfrac{s-4}{5}}}{x-1} \mathrm{d}x
\end{aligned}
$$
where you take $x^5\to x$. by recalling the representation of digamma function
$$
\psi(s+1) = -\gamma + \int_{0}^{1} {\frac{x^{s}-1}{x-1} \mathrm{d}x}
$$
we have
$$
I'(s)=\frac1{5}\left(\psi\left(\frac{s+1}{5}\right)+\psi\left(\frac{s+4}{5}\right)-\psi\left(\frac{s+2}{5}\right)-\psi\left(\frac{s+3}{5}\right)\right)
$$
with $\lim_{s\to\infty}I(s)=0$ and asymptotic expansion
$$\ln\Gamma(z) = (z-\tfrac1{2})\ln z - z + \ln2\pi + \tfrac1{12z} + o(z^{-3})$$
to anti-derivative
$$
I(0) = -\int_{0}^{\infty}I'(s)\,\mathrm{d}s = -\ln\left.\left(\frac{\Gamma\left(\frac{s+1}{5}\right)\Gamma\left(\frac{s+4}{5}\right)}{\Gamma\left(\frac{s+2}{5}\right)\Gamma\left(\frac{s+3}{5}\right)}\right)\right|_{s=0}^{\infty} = \ln\left(\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}\right)
$$
where using the reflection formula to obtain final answer
$$
\int_{0}^{1} \frac{x^2-1}{(x^4+x^3+x^2+x+1)\ln x} \mathrm{d}x = \ln\left(\frac{\sqrt5+1}{2}\right)
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[#ffd,15px]{\int_{0}^{1}{x^{2} - 1 \over \pars{x^{4} + x^{3} + x^{2} + x + 1}\ln\pars{x}}\,\dd x}
\\ = &\
\int_{0}^{1}{x^{2} - 1 \over
\pars{x^{5} - 1}/\pars{x - 1}}\ 
\overbrace{\pars{{1 \over x - 1}\int_{0}^{1}x^{y}\,\dd y}}
^{\ds{1 \over \ln\pars{x}}}\ \,\dd x =
\int_{0}^{1}\int_{0}^{1}{x^{y} - x^{y + 2} \over 1 - x^{5}}\,\dd x\,\dd y
\\[5mm] = &\
{1 \over 5}\int_{0}^{1}\int_{0}^{1}{x^{y/5 - 4/5} - x^{y/5 - 2/5} \over
1 - x}\,\dd x\,\dd y
\\[5mm] = &\
{1 \over 5}\int_{0}^{1}\pars{\int_{0}^{1}{1 - x^{y/5 - 2/5} \over
1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{y/5 - 4/5} \over
1 - x}\,\dd x}\dd y
\\[5mm] = &\
{1 \over 5}\int_{0}^{1}\bracks{\Psi\pars{{y \over 5} + {3 \over 5}} -
\Psi\pars{{y \over 5} + {1 \over 5}}}\dd y
\\[5mm] = &\
\bracks{\ln\pars{\Gamma\pars{{y \over 5} + {3 \over 5}}} -
\ln\pars{\Gamma\pars{{y \over 5} + {1 \over 5}}}}_{\ 0}^{\ 1}
\\[5mm] = &\
\ln\pars{\ln\pars{\Gamma\pars{4/5}}\ln\pars{\Gamma\pars{1/5}} \over \ln\pars{\Gamma\pars{2/5}}\ln\pars{\Gamma\pars{3/5}}} =
\ln\pars{\sin\pars{2\pi/5} \over \sin\pars{\pi/5}}
\\[5mm] = &\
\bbx{\ln\pars{2\cos\pars{\pi \over 5}}} =
\bbx{\ln\pars{1 + \root{5} \over 2}}\
\approx\ 0.4812
\end{align}
