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.

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, $\Gamma(z)\Gamma(1-z) = \pi /\sin (\pi z)$
$$ = \ln \frac{\sin (2\pi/5)}{\sin (\pi/5)}$$
$$ = \ln (2\cos (\pi/5))$$
$$ = \ln (\phi)$$
