<s>[INST] 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.
 [/INST] (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
$$
\int_{0}^{1} \frac{x^2-1}{(x^4+x^3+x^2+x+1)\ln x} \mathrm{d}x = \ln\left(\frac{\sqrt5+1}{2}\right)
$$
</s>