Integral from 0 to 1 ln(x)ln(1-x) dx

BPRP achieved this result using power series. Today I will solve it using the Bose Integral.

\begin{align*}
& \int_0^1 \ln{(x)}\ln{(1-x)} dx = 2 - \dfrac{\pi^2}{6} \\& \\&{\rm{Proof}}: \\& \\&\therefore \int_0^1 \ln{(x)}\ln{(1-x)} dx \\& \\&= \ln{(x)}(-x-\ln{(1-x)}+x\ln{(1-x)}) \mid_0^1 \\& \\&+ \int_0^1 1-\ln{(1-x)} dx + \int_0^1 \dfrac{\ln{(1-x)}}{x} dx \\& \\&Note: \int_0^1 \ln{(1-x)} dx = -1 \\& \\& = 0 + 2 + \int_0^1 \dfrac{\ln{(1-x)}}{x} dx \\& \\& Note: \int_0^1 \dfrac{\ln{(1-x)}}{x} dx = - \int_0^1 \dfrac{\ln{(x)}}{x-1} dx = - \dfrac{\pi^2}{6} \\& \\&\therefore \int_0^1 \ln{(x)}\ln{(1-x)} dx = 2 - \dfrac{\pi^2}{6} \Box\end{align*}

For proof of the last note: https://cdchester.co.uk/2019/10/18/integral-from-0-to-1-lnx-x-1-dx/

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