Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$

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, $\Gamma(z)\Gamma(1-z) = \pi /\sin (\pi z)$

$$ = \ln \frac{\sin (2\pi/5)}{\sin (\pi/5)}$$ $$ = \ln (2\cos (\pi/5))$$ $$ = \ln (\phi)$$

(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 $$ \int_{0}^{1} \frac{x^2-1}{(x^4+x^3+x^2+x+1)\ln x} \mathrm{d}x = \ln\left(\frac{\sqrt5+1}{2}\right) $$