Integral from 0 to 1 [ln(x)/(x-1)] dx

BPRP solved this integral using power series back in 2017. Today I will solve it using the Bose Integral.

\begin{align*}
& \int_0^1 \dfrac{\ln{(x)}}{x-1} dx = \dfrac{\pi^2}{6} \\& \\&Proof: \\& \\&\therefore u= \ln{(x)} \: , \: x = e^u \Longrightarrow
dx = e^u du \\& \\&\therefore \int_0^1 \dfrac{\ln{(x)}}{x-1} dx \Longrightarrow \int_{-{\infty}}^0 \dfrac{ue^u}{e^u-1} du \\& \\&\therefore u= -g \: \Longrightarrow
du = -dg \\& \\& \therefore \int_{-{\infty}}^0 \dfrac{ue^u}{e^u-1} du \Longrightarrow \int_0^\infty \dfrac{ge^{-g}}{1-e^{-g}} dg \\& \\&Note: \: \int_0^\infty \dfrac{ge^{-g}}{1-e^{-g}} dg = \int_0^\infty \dfrac{g}{e^{g}-1} dg \\& \\& Note: \: \int_0^\infty \dfrac{g^{n}}{e^{g}-1} dg = (n)!\zeta{(n+1)} \\& \\&\therefore \int_0^\infty \dfrac{g}{e^{g}-1} dg = 1!\zeta{(2)} = \dfrac{\pi^2}{6} \: \Box
\end{align*}

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