Question

$\displaystyle\tt \red{\int_{0}^{1} \frac{ {x}^{2} - 1}{( {x}^{4} + {x}^{3} + {x}^{2} + {x}^{} + 1) \ln(x) } \: dx}$​

Answer 1

$\displaystyle\tt \red{\int_{0}^{1} \frac{ {x}^{2} - 1}{( {x}^{4} + {x}^{3} + {x}^{2} + {x}^{} + 1) \ln(x) } \: dx}$

$\displaystyle\tt \red{\int_{0}^{1} \frac{ {x}^{2} - 1}{ \left[ \bigg(\frac{{x}^{5} - 1}{x - 1} \bigg) \right] \ln(x) } \: dx}$

$\displaystyle\tt \red{\int_{0}^{1} \frac{ - {x}^{3} + {x}^{2} + x - 1}{(1 - {x}^{5}) \ln(x)}\: dx }$

$\displaystyle\tt \red{\int_{0}^{1} \frac{ { - x}^{ \frac{ - 1}{5} } + {x}^{ \frac{ - 2}{5} } + {x}^{ \frac{ - 3}{5} } - {x}^{ \frac{ - 4}{5} } }{1 - x} \: \frac{1}{ \ln(x) } \: dx }$

$\displaystyle\tt \red{{\int_{0}^{ \infty }\int_{0}^{1} \frac{ {x}^{t - \frac{1}{5} } - {x}^{t - \frac{2}{5} } - {x}^{t - \frac{3}{5} } + {x}^{t - \frac{4}{5} } }{1 - x} } \: dx \: dt}$

$\displaystyle \tt \red{\int_{0}^{ \infty } \bigg( - \int_{0}^{1} \frac{1 - {x}^{t - \frac{1}{5} } }{1 - x} dx + \int_{0}^{1}\frac{1 - {x}^{t - \frac{2}{5} } }{1 - x}dx +\int_{0}^{1}\frac{1 - {x}^{t - \frac{3}{5} } }{1 - x}dx - \int_{0}^{1}\frac{1 - {x}^{t - \frac{4}{5} } }{1 - x}dx \bigg)dt}$

$\displaystyle \tt \red {\int_{0}^{ \infty } \bigg[ - \Psi \bigg( t + \frac{4}{5} \bigg) + \Psi \bigg(t + \frac{3}{5} \bigg) + \Psi \bigg(t + \frac{2}{5} \bigg) - \Psi \bigg(t + \frac{1}{5} \bigg)\bigg] \: dt}$

$\tt \red{ \ln \bigg( \dfrac{ \Gamma(t + \frac{3}{5} ) \Gamma(t + \frac{2}{5} )}{ \Gamma(t + \frac{4}{5}) \Gamma(t + \frac{1}{5} ) } \bigg) \bigg |^{t \to \infty }_{t = 0} }$

$\tt \red{ \ln(1) - \ln \bigg( \dfrac{ \Gamma( \frac{3}{5} ) \Gamma(\frac{2}{5} )}{ \Gamma(\frac{4}{5}) \Gamma( \frac{1}{5} ) } \bigg) }$

$\tt \red{ \ln \bigg( \dfrac{ \frac{\pi}{ \sin } ( \frac{\pi}{5} )}{ \frac{\pi}{ \sin} ( \frac{2\pi}{5} )} \bigg) }$

$\tt \red{ \ln(2 \cos( \frac{\pi}{5} ) ) }$

$\red{ \ln( \phi) }$