$ \int^1_0 x^m(\ln x)^n ~dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}} , \quad m>-1,\, n=0,1,2,\dots \\ $ $ \int^1_0 \dfrac{\ln x}{1+x} ~dx = -\dfrac{\pi^2}{12} \\$ $ \int^1_0 \dfrac{\ln x}{1-x}~dx = -\dfrac{\pi^2}{6} \\$ $ \int^1_0 \dfrac{\ln (1+x)}{x}~ dx = \dfrac{\pi^2}{12} \\ $ $  \int^1_0 \dfrac{\ln (1 - x)}{x}~ dx = - \dfrac{\pi^2}{6} \\ $ $ \int^1_0 \ln x \ln (1+x) \,dx = 2 - 2\ln 2 - \dfrac{\pi^2}{12} \\ $ $ \int^1_0 \ln x \ln (1 - x) \,dx = 2 - \dfrac{\pi^2}{6} \\ $ $ \int^\infty_0 \dfrac{x^{p-1}\,\ln x}{1+x}~ dx = -\pi^2\, \csc (p\pi)\,\cot (p\pi), 0 < p < 1 \\ $ $ \int^1_0 \dfrac{x^m - x^n}{\ln x} dx = \ln \dfrac{m+1}{n+1} \\ $ $ \int^\infty_0 e^{-x}\,\ln x\,dx = - \gamma \\ $ $ \int^\infty_0 e^{-x^2} \ln x \, dx = -\dfrac{\sqrt{\pi}}{4} ( \gamma + 2\,\ln 2 ) \\ $ $ \int^\infty_0 \ln \left(\dfrac{e^x+1}{e^x-1} \right) dx = \dfrac{\pi^2}{4} \\ $ $ \int^{\pi/2}_0 \ln (\sin x)~ dx = \int^{\pi/2}_0 \ln (\cos x) dx = -\dfrac{\pi}{2}~ \ln 2 \\ $ $ \int^{\pi/2}_0 ( \ln (\sin x))^2 ~dx = \int^{\pi/2}_0 ( \ln (\cos x))^2 dx = \dfrac{\pi}{2}~(\ln 2)^2 + \dfrac{\pi^3}{24} \\ $ $ \int^\pi_0 x\,\ln(\sin x)~ dx = -\dfrac{\pi^2}{2}\,\ln 2 \\ $ $ \int^{\pi/2}_0 \sin x \ln (\sin x) ~dx = \ln 2 - 1 \\ $ $ \int^{2\pi}_0 \ln (a + b \sin x)~ dx = \int^{2\pi}_0 \ln (a + b \cos x) dx = 2\pi \ln\left(a + \sqrt{a^2-b^2} \right) \\$ $ \int^\pi_0 \ln (a + b\cos x) dx = \pi \ln \left( \dfrac{a + \sqrt{a^2 - b^2}}{2} \right) \\ $ $ \int^\pi_0 \ln\left( a^2 - 2ab \cos x + b^2 \right) dx =              \begin{cases} 2\pi\ln a & a \geq b > 0 \\ 2\pi \ln b & b \geq a > 0 \end{cases} \\ $ $ \int^{\pi/4}_0 \ln(1 + \tan x)dx= \dfrac{\pi}{8} ~\ln 2 \\ $ $ \int^\dfrac{\pi}{2}_0  \sec x \,\ln\left(\dfrac{1+b\cos x}{1 + a \cos x}\right) dx = \dfrac{1}{2}\left(\arccos^2 a - \arccos^2b\right) \\ $         

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