$$ \begin {aligned} \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {dxdydz}{xyz-xy-yz-zx} &=\int _{0}^{1}\int _0^{1}\left .\frac {\ln \left (-xyz+xy+yz+zx\right )}{xy-x-y}\right|_{z=0}^1dxdy\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln \left (x+y\right )-\ln \left (xy\right )}{1-\left (1-x\right )\left (1-y\right )}dxdy\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln (2-x-y)}{1-xy}dxdy+\int _{0}^{1}\int _{0}^{1}\frac {\ln \left (1-x\right )+\ln \left (1-y\right )}{1-xy}dxdy\\ &=-2\iint _{0< y< x< 1}\frac {\ln \left (2-x-y\right )}{1-xy}dxdy+2\int _{0}^{1}\int _{0}^{1}\frac {\ln \left (1-x\right )}{1-xy}dydx\\ &=-\left (\int _{0}^{1}\int _{0}^{s}+\int _{1}^{2}\int _{0}^{2-s}\right )\frac {\ln (2-s)}{1-\frac {s^{2}-t^{2}}4}dtds-2\int _{0}^{1}\frac {\ln ^{2}\left (1-x\right )}xdx\quad \left (s=x+y,t=x-y\right )\\ &=-4\underbrace{\int _{0}^{1}\int _{0}^{s}\frac {\ln \left (2-s\right )}{4-s^2+t^2}dtds}_{I_1}-4\underbrace{\int _{1}^{2}\int _{0}^{2-s}\frac {\ln \left (2-s\right )}{4-s^2+t^2}dtds}_{I_2}-4\zeta (3) \end {aligned} \\ \begin {aligned} I_1&=\int _{0}^{1}\int _{0}^{s}\frac {\ln \left (2-s\right )}{4-s^2+t^2}dtds\\ &=\int _{0}^{1}\frac {\ln (2-s)}{\sqrt {4-s^{2}}}\arctan \frac {s}{\sqrt {4-s^{2}}}ds\\ &=\int _{0}^{\frac {1}2}\frac {\ln \left (2\left (1-s\right )\right )}{\sqrt {1-s^2}}\arctan \frac {s}{\sqrt {1-s^{2}}}ds\\ &=\int _{0}^{\frac {\pi }6}x\ln \left (2\left (1-\sin x\right )\right )dx\\ &=\int _{0}^{\frac {\pi }6}x\ln \left (4\cdot \frac {1-\cos \left (\frac {\pi }2-x\right )}2\right )dx\\ &=2\int _{0}^{\frac {\pi }6}x\ln \left (2\sin \frac {\frac {\pi }2-x}2\right )dx\\ &=-2\sum _{n=1}^\infty \frac {1}n\int _{0}^{\frac {\pi }6}x\cos \left (\frac {\pi n}2-nx\right )dx\\ &=-2\sum _{n=1}^\infty \frac {1}{n^{3}}\int _{0}^{\frac {n\pi }6}x\cos \left (x-\frac {\pi n}2\right )dx\\ &=-2\sum _{n=1}^\infty \frac {1}{n^{3}}\left (\left .x\sin \left (x-\frac {\pi n}2\right )\right|_0^{\frac {n\pi }6}-\int _{0}^{\frac {n\pi }6}\sin \left (x-\frac {\pi n}2\right )dx\right )\\ &=-2\sum _{n=1}^\infty \left (\frac {\pi }{6n^{2}}\sin \left (\frac {n\pi }6-\frac {n\pi }2\right )+\frac {\cos \left (\frac {n\pi }6-\frac {n\pi }2\right ) -\cos \frac {n\pi }2}{n^3}\right )\\ &=\frac {\pi }3\Im \operatorname {Li} _2 \left (e^{\frac {\pi i}3}\right )-2\Re \operatorname{Li} _{3} \left (e^{\frac {\pi i}3}\right )+2\Re \operatorname{Li} _{3} \left (i\right ) \end {aligned} \\ \begin {aligned} I_2&=\int _{1}^{2}\int _{0}^{2-s}\frac {\ln \left (2-s\right )}{4-s^2+t^2}dtds\\ &=\int _{1}^{2}\frac {\ln \left (2-s\right )}{\sqrt {4-s^{2}}}\arctan \frac {2-s}{\sqrt {4-s^2}}ds\\ &=\int _{\frac {1}2}^{1}\frac {\ln \left (2\left (1-s\right )\right )}{\sqrt {1-s^2}}\arctan \sqrt {\frac {1-s}{1+s}}ds\\ &=\int _0^{\frac {\pi }3}\frac {x}2\ln \left (2\left (1-\cos x\right )\right )dx\\ &=\int _{0}^{\frac {\pi }3}x\ln \left (2\sin \frac {x}2\right )\\ &=-\sum _{n=1}^\infty \frac {1}n\int _{0}^{\frac {\pi }3}x\cos (nx)dx\\ &=-\sum _{n=1}^\infty \frac {1}{n^3}\int _{0}^{\frac {n\pi }3}x\cos xdx\\ &=-\sum _{n=1}^\infty \frac {1}{n^{3}}\left (\left .x\sin x\right|_0^{\frac {n\pi }3}-\int _{0}^{\frac {n\pi }3}\sin xdx\right )\\ &=-\sum _{n=1}^\infty \left (\frac {\pi \sin \frac {n\pi }3}{3n^2}+\frac {\cos \frac {n\pi }3-1}{n^{3}}\right )\\ &=-\frac {\pi }3\Im \operatorname{Li} _{2} \left (e^{\frac {\pi i}3}\right )-\Re \operatorname {Li} _3 \left (e^{\frac {\pi i}3}\right )+\zeta (3) \end {aligned} \\ \begin {aligned} \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {dxdydz}{xyz-xy-yz-zx} &=-4I_1-4I_2-4\zeta (3)\\ &=-4\left (\frac {\pi }3\Im \operatorname{Li} _{2} \left (e^{\frac {\pi i}3}\right )-2\Re \operatorname{Li} _{3} \left (e^{\frac {\pi i}3}\right )+2\Re \operatorname{Li} _{3} (i)\right )\\ &\quad -4\left (-\frac {\pi }3\Im \operatorname{Li} _{2} \left (e^{\frac {\pi i}3}\right )-\Re \operatorname{Li} _{3} \left (e^{\frac {\pi i}3}\right )+\zeta (3)\right )-4\zeta (3) \\ &=12\Re \operatorname{Li} _{3} \left (e^{\frac {\pi i}3}\right )-8\Re \operatorname{Li} _{3} (i)-4\zeta (3)-4\zeta (3)\\ &=4\zeta (3)+\frac {3}4\zeta (3)-4\zeta (3)-4\zeta (3)\\ &=\textcolor {blue}{-\frac {13}{4}\zeta (3)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {dxdydz}{xyz-x-y-z}\\ &=\int _{0}^{1}\int _{0}^{1}\left .\frac {\ln (-xyz+x+y+z)}{xy-1}\right |_{z=0}^1dxdy\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln (1+x+y-xy)}{1-xy}+\int _{0}^{1}\int _{0}^{1}\frac {\ln (x+y)}{1-xy}dxdy\\ &=-\frac {13}{16}\zeta (3)+\frac {1}8\zeta (3)\\ &=\textcolor {blue}{-\frac {11}{16}\zeta (3)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\frac {\ln \left (1+x+y-xy\right )}{1-xy}dxdy \\ &=\int _{0}^{1}\int _{0}^{1}\frac {\ln \left (2-(1-x)(1-y)\right )}{1-xy}dxdy\\ &=\int _{0}^{1}\int _{0}^{1}\frac {\ln \left (2-xy\right )}{1-(1-x)(1-y)}dxdy\\ &=\int _{0}^{1}\int _{0}^{y}\frac {\ln \left (2-x\right )}{1-\left (1-\frac {x}y\right )\left (1-y\right )}\frac {dx}ydy\\ &=\int _{0< x< y< 1}\frac {\ln (2-x)}{x-xy+y^2}dxdy\\ &=\int _{0}^{1}\ln (2-x)\int _{x}^{1}\frac {dydx}{x-xy+y^2}\\ &=2\int _{0}^{\frac {1}2}\ln (2-2x)\int _{2x}^{1}\frac {dydx}{2x-2xy+y^2}\\ &=2\int _{0}^{\frac {1}2}\ln \left (2(1-x)\right )\int _{2x}^{1}\frac {dydx}{(y-x)^2+1-(1-x)^2}\\ &=2\int _{\frac {1}2}^{1}\ln \left (2x\right )\int _{2(1-x)}^{1}\frac {dydx}{\left (y-(1-x)\right )^2+1-x^2}\\ &=2\int _{\frac {1}2}^{1}\frac {\ln \left (2x\right )}{\sqrt {1-x^{2}}}\left .\arctan \frac {y-(1-x)}{\sqrt {1-x^{2}}}\right|_{y=2(x-1)}^1dx\\ &=2\int _{0}^{\frac {\pi }3}\ln \left (2\cos \theta \right )\left (\arctan \frac {\cos \theta }{\sqrt {1-\cos ^{2}\theta }}-\arctan \frac {1-\cos \theta }{\sqrt {1-\cos ^{2}\theta }}\right )d\theta \\ &=2\int _{0}^{\frac {\pi }3}\ln \left (2\cos \theta \right )\left (\frac {\pi }2-\theta -\frac {\theta }2\right )d\theta \\ &=-2\int _{0}^{\frac {\pi }3}\sum _{n=1}^\infty \frac {(-1)^n\cos 2n\theta }n\left (\frac {\pi }2-\frac {3\theta }2\right )d\theta \\ &=-2\sum _{n=1}^\infty \frac {(-1)^n}{n}\int _{0}^{\frac {\pi }3}\cos \left (2n\theta \right )\left (\frac {\pi }2-\frac {3\theta }2\right )d\theta \\ &=-\sum _{n=1}^\infty \frac {(-1)^n}{n^2}\int _{0}^{\frac {2n\pi }3}\cos (\theta )\left (\frac {\pi }2-\frac {3\theta }{4n}\right )d\theta \\ &=-\sum _{n=1}^\infty \frac {(-1)^n}{n^2}\left (\left (\frac {\pi }2-\frac {\pi }{2}\right )\sin \frac {2n\pi }3+\frac {3}{4n}\int _{0}^{\frac {2n\pi }3}\sin \theta d\theta \right )\\ &=-\frac {3}4\sum _{n=1}^\infty \frac {(-1)^n}{n^3}\left (1-\cos \frac {2n\pi }3\right )\\ &=-\frac {3}4\left (\operatorname{Li} _{3} (-1)-\Re \operatorname{Li} _{3} \left (\frac {1-\sqrt 3i}2\right )\right )\\ &=-\frac {3}4\left (-\frac {3}4\zeta (3)-\frac {1}3\zeta (3)\right )\\ &=\textcolor {blue}{\frac {13}{16}\zeta (3)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\frac {\ln (1+x+y-xy)}{1-(1-x)(1-y)}dxdy\\ &=\int _{0}^{1}\int _{0}^{1}\frac {\ln (2-xy)}{1-xy}dxdy\\ &=\iint _{0< x< y< 1}\frac {\ln (2-x)}{1-x}\frac {dx}ydy\\ &=-\int _{0}^{1}\frac {\ln x\ln (2-x)}{1-x}dx\\ &=-\int _{0}^{1}\frac {\ln (1-x)\ln (1+x)}xdx\\ &=-\sum _{0< n,m}\int _{0}^{1}\frac {x^n(-x)^m}{nm}\frac {dx}x\\ &=-\sum _{0< n,m}\frac {(-1)^{m}}{nm(n+m)}\\ &=-\sum _{0< n,m}\left (\frac {(-1)^n(-1)^{n+m}}{n(n+m)^2}+\frac {(-1)^{m}}{m(n+m)^{2}}\right )\\ &=-\zeta \left (\bar {1},\bar {2}\right )-\zeta \left (\bar {1},2\right )\\ &=-\frac {\pi ^{2}\ln 2}4+\frac {13}8\zeta (3)-\zeta (3)+\frac {\pi ^{2}\ln 2}4\\ &=\textcolor {blue}{\frac {5}8\zeta (3)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln x}{1-x(1-y(1-z))}dxdydz\\ &=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln x}{1-x(1-yz)}dxdydz\\ &=\int _{0}^{1}\ln x\iint _{0< y< z< 1}\frac {1}{1-x(1-y)}\frac {dy}zdzdx\\ &=\int _{0}^{1}\ln x\int _{0}^{1}\frac {-\ln y}{1-x(1-y)}dydx\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln x\ln y}{1-x(1-y)}dxdy\\ &=\int _{0}^{1}\frac {\ln y}{1-y}\left [\operatorname{Li} _{2} \left (x(1-y)\right )+\ln x\ln \left (1-x(1-y)\right )\right ]_{x=0}^1dy\\ &=\int _{0}^{1}\frac {\ln y\operatorname{Li} _{2} (1-y)}{1-y}dy\\ &=\int _{0}^{1}\frac {\ln (1-x)\operatorname{Li} _{2}(x) }xdx\\ &=-\sum _{0< n,m}\int _{0}^{1}\frac {x^{n}x^n}{nm^2}\frac {dx}x\\ &=-\sum _{0< n,m}\frac {1}{nm^{2}(n+m)}\\ &=-\frac {1}2\sum _{0< n,m}\left (\frac {1}{nm^{2}(n+m)}+\frac {1}{n^{2}m(n+m)}\right )\\ &=-\frac {1}2\sum _{0 < n,m}\frac {1}{n^{2}m^{2}}\\ &=-\frac {1}2\zeta (2)^{2}\\ &=\textcolor {blue}{-\frac {5}4\zeta (4)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln y}{1-x(1-y(1-z))}dxdydz\\ &=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln y}{1-x(1-yz)}dxdydz\\ &=-\int _{0}^{1}\left .\int _{0}^{1}\frac {\ln y\ln (1-x(1-yz))}{1-yz}\right |_{x=0}^1dydz\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln y\ln yz}{1-yz}dydz\\ &=-\iint _{0< z< y< 1}\frac {\ln y\ln z}{1-z}\frac {dz}ydy\\ &=-\int _{0}^{1}\left .\frac {\ln z}{1-z}\frac {\ln ^{2}y}2\right|_{y=z}^1\\ &=\frac {1}2\int _{0}^{1}\frac {\ln ^{3}z}{1-z}dz\\ &=\frac {(-1)^33!}2\zeta (4)\\ &=\textcolor {blue}{-3\zeta (4)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln z}{1-x(1-y(1-z))}dxdydz\\ &=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-z)}{1-x(1-yz)}dxdydz\\ &=-\int _{0}^{1}\int _{0}^{1}\left .\frac {\ln (1-z)}{1-yz}\ln (1-x(1-yz))\right|_{x=0}^1dydz\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-z)\ln yz}{1-yz}dydz\\ &=-\iint _{0< y< z< 1}\frac {\ln (1-z)\ln y}{1-y}\frac {dy}zdz\\ &=\int _{0}^{1}\left .\frac {\ln y}{1-y}\operatorname{Li} _{2}\left (z\right ) \right|_{z=y}^1dz\\ &=\zeta (2)\int _{0}^{1}\frac {\ln y}{1-y}dy-\int _{0}^{1}\frac {\ln y\operatorname{Li} _{2} (y)}{1-y}dy\\ &=-\zeta (2)^{2}-\sum _{0< n,m}\frac {1}{n^{2}}\int _{0}^{1}y^{n+m}\ln y\frac {dy}y\\ &=-\frac {5}2\zeta (4)+\sum _{0< n,m}\frac {1}{n^2(n+m)^2}\\ &=-\frac {5}2\zeta (4)+\zeta (2,2)\\ &=-\frac {5}2\zeta (4)+\frac {3}4\zeta (4)\\ &=\textcolor {blue}{-\frac {7}4\zeta (4)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-x)}{1-x(1-y(1-z))}dxdydz\\ & =\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-x)}{1-x(1-yz)}dxdydz\\ &=\int _{0}^{1}\iint _{0< y< z< 1}\frac {\ln (1-x)}{1-x(1-y)}\frac {dy}zdzdx\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-x)\ln y}{1-x(1-y)}dydx\\ &=-\int _{0}^{1}\frac {\ln (1-x)}x\int _{0}^{1}\frac {\ln y}{y-\frac {x-1}x}dydx\\ &=-\int _{0}^{1}\left .\frac {\ln (1-x)}x\left (\ln y\ln \left (1-\frac {xy}{x-1}\right )+\operatorname{Li} _{2}\left (\frac {xy}{x-1}\right ) \right )\right|_{y=0}^1dx\\ &=-\int _{0}^{1}\frac {\ln (1-x)}x\operatorname{Li} _{2}\left (1-\frac {1}{1-x}\right ) dx\\ &=\int _{0}^{1}\frac {\ln (1-x)}x\left (\frac {1}2\ln ^{2}(1-x)+\operatorname{Li} _{2} (x)\right )dx\\ &=\frac {1}2\int _{0}^{1}\frac {\ln ^{3}(1-x)}xdx+\int _{0}^{1}\frac {\ln (1-x)\operatorname{Li} _{2} (x)}xdx\\ &=-3\zeta (4)-\sum _{0< n,m}\frac {1}{nm^{2}}\int _{0}^{1}x^{n+m}\frac {dx}x\\ &=-3\zeta (4)-\sum _{0< n,m}\frac {1}{nm^2(n+m)}\\ &=-3\zeta (4)-\frac {1}2\sum _{0< n,m}\left (\frac {1}{nm^{2}(n+m)}+\frac {1}{n^2m(n+m)}\right )\\ &=-3\zeta (4)-\frac {1}2\sum _{0< n,m}\frac {1}{n^{2}m^{2}}\\ &=-3\zeta (4)-\frac {1}2\zeta (2)^{2}\\ &=-3\zeta (4)-\frac {5}4\zeta (4)\\ &=\textcolor {blue}{-\frac {17}4\zeta (4)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-y)}{1-x(1-y(1-z))}dxdydz\\ &=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-y)}{1-x(1-yz)}dxdydz\\ &=-\int _{0}^{1}\int _{0}^{1}\left .\frac {\ln (1-y)}{1-xy}\ln (1-x(1-yz))\right|_{x=0}^1dydz\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-y)\ln (yz)}{1-yz}dydz\\ &=-\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-z)\ln yz}{1-yz}dydz\\ &=-\iint _{0< y< z< 1}\frac {\ln (1-z)\ln y}{1-y}\frac {dy}zdz\\ &=\int _{0}^{1}\left .\frac {\ln y}{1-y}\operatorname{Li} _{2}\left (z\right ) \right|_{z=y}^1dz\\ &=\zeta (2)\int _{0}^{1}\frac {\ln y}{1-y}dy-\int _{0}^{1}\frac {\ln y\operatorname{Li} _{2} (y)}{1-y}dy\\ &=-\zeta (2)^{2}-\sum _{0< n,m}\frac {1}{n^{2}}\int _{0}^{1}y^{n+m}\ln y\frac {dy}y\\ &=-\frac {5}2\zeta (4)+\sum _{0< n,m}\frac {1}{n^2(n+m)^2}\\ &=-\frac {5}2\zeta (4)+\zeta (2,2)\\ &=-\frac {5}2\zeta (4)+\frac {3}4\zeta (4)\\ &=\textcolor {blue}{-\frac {7}4\zeta (4)}. \end {aligned} $$

$$ \begin {aligned} &\int _0^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-zx)}{1-x(1-y(1-z))}dxdydz\\ &=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-zx)}{1-x+yx(1-z)}dydxdz\\ &=\int _{0}^{1}\int _{0}^{1}\left .\frac {\ln (1-zx)}{x(1-z)}\ln \left (1-x+yx(1-z)\right )\right|_{y=0}^1dxdz\\ &=\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-zx)}{x(1-z)}\left (\ln (1-zx)-\ln (1-x)\right )dxdz\\ &=\iint _{0< z< x< 1}\frac {\ln (1-z)}{x(x-z)}\left (\ln (1-z)-\ln (1-x)\right )dxdz\\ &=-\int _{0< x< z< 1}\frac {\ln z\ln \frac {z}x}{(1-x)(x-z)}dxdz\\ &=-\int _{0}^{1}\int _{1}^{\frac {1}x}\frac {\ln (xz)\ln z}{(1-x)(x-xz)}xdzdx\\ &=-\int _{0}^{1}\int _{x}^{1}\frac {\ln \frac {x}z\ln z}{(1-x)(1-z)z}dzdx\\ &=-\int _{0}^{1}\frac {\ln z}{(1-z)z}\int _{0}^{z}\frac {\ln \frac {x}z}{1-x}dxdz\\ &=-\int _{0}^{1}\frac {\ln z}{(1-z)z}\int _{0}^{1}\frac {\ln x}{\frac {1}z-x}dxdz\\ &=-\int _{0}^{1}\left .\frac {\ln z}{(1-z)z}\left (-\ln (x)\ln \left (1-zx\right )-\operatorname{Li} _{2} (zx)\right )\right |_{x=0}^1dz\\ &=\sum _{n=0}^\infty \int _{0}^{1}z^{n-1}\ln z\sum _{m=1}^\infty \frac {z^{m}}{m^{2}}dz\\ &=-\sum _{n=0}^\infty \sum _{m=1}^\infty \frac {1}{m^2(n+m)^2}\\ &=-\zeta ^{\star}(2,2)\\ &=-\zeta (2,2)-\zeta (4)\\ &=-\frac {3}4\zeta (4)-\zeta (4)\\ &=\textcolor {blue}{-\frac {7}4\zeta (4)}. \end {aligned} $$

$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {\ln (1-xyz)}{1-x(1-y(1-z))}dxdydz\\ &=\int _{0}^{1}\int _{0}^{1}\int _{0}^{xy}\frac {\ln (1-z)}{1-x+xy-z}\frac {dz}{xy}dxdy\\ &=\int _{0}^{1}\int _{0}^{x}\int _{0}^{y}\frac {\ln (1-z)}{1-x+y-z}\frac {dz}y\frac {dy}xdx\\ &=\iiint _{0< z< y< x< 1}\frac {\ln (1-z)}{y(1+y-z)}\left (\frac {1}{1-x+y-z}+\frac {1}x\right )dxdydz\\ &=\iint _{0< z< y< 1}\frac {\ln (1-z)}{y(1+y-z)}\left .\ln \frac {x}{1-x+y-z}\right|_{x=y}^1dydz\\ &=\iint _{0< z< y< 1}\frac {\ln (1-z)}{y(1+y-z)}\ln \frac {1-z}{y(y-z)}dydz\\ &=\iint _{0< y,z< 1}\frac {\ln (1-yz)}{1+y(1-z)}\ln \frac {1-yz}{y^2(1-z)}dydz\\ &=\iint _{0< y< z< 1}\frac {\ln (1-y)}{1-(1-y)(1-z)}\ln \frac {z^{2}(1-y)}{y^{2}(1-z)}\\ &=\iint _{0< z< y< 1}\frac {\ln y}{1-yz}\ln \frac {(1-z)^{2}y}{(1-y)^2z}dydz\\ &=\iiint _{0< z< x< y< 1}\frac {\ln y}{1-yz}\frac {dx}xdydz+2\iiint _{0< z< x< y}\frac {\ln y}{1-yz}\frac {dx}{1-x}dydz\\ &=-\iint _{0< x< y< 1}\left .\frac {\ln y}{xy}\ln (1-yz)\right |_{z=0}^xdxdy-2\iint _{0< x< y< 1}\left .\frac {\ln y}{y(1-x)}\ln (1-yz)\right |_{z=0}^xdxdy\\ &=-\iint _{0< x< y< 1}\frac {\ln y\ln (1-xy)}{xy}dxdy-2\iint _{0< x< y< 1}\frac {\ln y\ln (1-xy)}{y(1-x)}dxdy\\ &=\int _{0}^{1}\left .\frac {\ln y}y\operatorname{Li} _{2} (xy)\right|_{x=0}^ydy-2\int _{0}^{1}\left .\frac {1}{1-x}\left (-\ln y\operatorname {Li} _2(xy)+\operatorname {Li} _3(xy) \right )\right |_{y=x}^1dx\\ &=\int _{0}^{1}\frac {\ln y\operatorname {Li} _2 \left (y^{2}\right )}ydy-2\int _{0}^{1}\frac {\operatorname{Li} _{3} (x)+\ln x\operatorname {Li} _2 \left (x^{2}\right )-\operatorname{Li} _{3}\left (x^{2}\right ) }{1-x}\\ &=\frac {1}4\int _{0}^{1}\frac {\ln y\operatorname {Li} _2 (y)}ydy+2\left .\ln (1-x)\left (\operatorname{Li} _{3}(x)+\ln x\operatorname {Li} _2 \left (x^{2}\right ) -\operatorname{Li} _{3}\left (x^{2}\right ) \right )\right|_{x=0}^1-2\int _{0}^{1}\frac {\ln (1-x)}x\left (\operatorname{Li} _{2}(x)-\operatorname {Li} _2 \left (x^{2}\right ) -2\ln x\ln \left (1-x^{2}\right )\right )\\ &=\left .\frac {1}4\left (\ln y\operatorname{Li} _{3}(y) -\operatorname{Li} _{4}(x) \right )\right|_0^1 + \left .2\operatorname {Li} _2 (x)\left (\operatorname {Li} _2 (x)-\operatorname {Li} _2 \left (x^{2}\right )-2\ln x\ln \left (1-x^{2}\right )\right )\right|_0^1-2\int _{0}^{1}\operatorname {Li} _2 (x)\left (\frac {-\ln (1-x)}x+\frac {2x\ln \left (1-x^{2}\right )}{x^2}-\frac {2\ln \left (1-x^{2}\right )}x+\frac {4x\ln x}{1-x^2}\right )dx\\ &=-\frac {1}4\zeta (4)-2\int _{0}^{1}\operatorname {Li} _2 (x)\left (\frac {-\ln (1-x)}x+\frac {2\ln x}{1-x}-\frac {2\ln x}{1+x}\right )dx\\ &=-\frac {1}4\zeta (4)-2\sum _{0< n,m}\int _{0}^{1}\left (\frac {x^{n+m-1}}{n^{2}m}+\frac {2\ln (x)x^{n+m-1}}{n^{2}}+\frac {2\ln (x)(-1)^mx^{n+m-1}}{n^2}\right )dx\\ &=-\frac {1}4\zeta (4)-2\sum_{0< n,m} \frac {1}{n^2m(n+m)}+4\sum _{0< n,m}\frac {1}{n^2(n+m)^2}+4\sum _{0< n,m}\frac {(-1)^{n}(-1)^{n+m}}{n^2(n+m)^2}\\ &=-\frac 14\zeta (4)-\sum _{0< n,m}\left (\frac {1}{n^{2}m(n+m)}+\frac {1}{nm^{2}(n+m)}\right )+4\zeta (2,2)+4\zeta (\bar {2},\bar {2})\\ &=-\frac {1}4\zeta (4)-\zeta (2)^{2}+3\zeta (4)-\frac {3}4\zeta (4)\\ &=\textcolor {blue}{-\frac {1}2\zeta (4)}. \end {aligned} $$

$$ \begin{align*} &\iiint_{0< x,y,z< 1}\frac{\log(1-x(1-y))}{1-x(1-y(1-z))}dxdydz\\ =&\iiint_{0< z< y< x< 1}\frac{\log(1-x+y)}{1-x+y-z}\frac1{xy}dxdydz\\ =&\iiint_{0< z< y< x< 1}\frac{\log x}{x-z}\frac1{y(1-x+y)}dxdydz\\ =&-\iint_{0< z< x< 1}\frac{\log x}{x-z}\left[\frac1{1-x}\log\left(\frac{1-x+y}y\right)\right]_z^xdxdz\\ =&-\iint_{0< z< x< 1}\frac{\log x}{x-z}\frac1x\log\left(\frac z{x(1-x+z)}\right) dxdz\\ =&-\iint_{0< x,z< 1}\frac{\log x}{1-x}\frac1{1-x}\log\left(\frac z{1-x(1-z)}\right) dxdz\\ =&-\iint_{0< x,z< 1}\frac{\log x\log z}{(1-x)(1-z)} dxdz+\iint_{0< x,z< 1}\frac{\log x\log (1-xz)}{z(1-x)} dxdz\\ =&-\left(\left[\text{Li}_2(1-x)\right]_0^1\right)^2-\sum_{0< n,m}\frac1m\int_0^1x^{n+m-1}\log xdx\int_0^1 z^{m-1}dz\\ =&-\zeta(2)^2+\sum_{0< n,m}\frac1{m^2(n+m)^2}\\ =&-\frac52\zeta(4)+\zeta(2,2)\\ =&-\frac52\zeta(4)+\frac34\zeta(4)\\ =&\textcolor {blue}{-\frac74\zeta(4)}. \end{align*} $$

$$ \begin{align*} &\iiint_{0< x,y,z< 1}\frac{\log(1-x(1-z))}{1-x(1-y(1-z))} dxdydz\\ =&\iiint_{0< x,y,z< 1}\frac{\log(1-xz)}{1-x+xyz} dxdydz\\ =&\iiint_{0< y< z< x< 1}\frac{\log(1-z)}{1-x+y}\frac1{xz}dxdydz\\ =&-\iint_{0< y< x< 1}\frac{\text{Li}_2(x)-\text{Li}_2(y)}{x(1-x+y)} dxdy\\ =&\int_0^1\frac{\text{Li}_2(y)}{1+y}\left[\log\frac x{1-x+y}\right]_y^1dy-\int_0^1\frac{\text{Li}_2(x)}x\left[\log(1-x+y)\right]_0^x dx\\ =&-2\int_0^1\frac{\text{Li}_2(y)\log y}{1+y}dy+\int_0^1\frac{\text{Li}_2(x)\log(1-x)}xdx\\ =&-2\sum_{0< n,m}\frac{(-1)^{m-1}}{n^2}\int_0^1 y^{n+m-1}\log ydy-\sum_{0< n,m}\frac1{n^2}\int_0^1 x^{n+m-1} dx\\ =&-2\sum_{0< n,m}\frac{(-1)^m}{n^2(n+m)^2}-\sum_{0< n,m}\frac1{n^2m(n+m)}\\ =&-2\zeta(\bar2,\bar2)-\frac12\zeta(2)^2\\ =&\frac38\zeta(4)-\frac54\zeta(4)\\ =&\textcolor{blue}{-\frac78\zeta(4)}. \end{align*} $$

$$ \begin {aligned} &\iiint _{0< x,y,z< 1}\frac {\ln (1-y(1-z))}{1-x(1-y(1-z))}dxdydz\\ &=\iiint _{0< x,y,z< 1}\frac {\ln (1-yz)}{1-x(1-yz)}dxdydz\\ &=-\iint _{0< y,z< 1}\left .\frac {\ln (1-yz)}{1-yz}\ln (1-x(1-yz))\right|_{x=0}^1dydz\\ &=-\iint _{0< y,z< 1}\frac {\ln (1-yz)\ln (yz)}{1-yz}dydz\\ &=-\iint _{0< z< y< 1}\frac {\ln (1-z)\ln z}{1-z}\frac {dz}ydy\\ &=\int _{0}^{1}\frac {\ln (1-z)\ln ^{2}z}{1-z}dz\\ &=-\sum _{0< n,m}\frac {1}{n}\int _{0}^{1}z^{n+m-1}\ln ^{2}zdz\\ &=-2\sum _{0< n,m}\frac {1}{n(n+m)^3}\\ &=-2\zeta (1,3)\\ &=\textcolor {blue}{-\frac {1}2\zeta (4)}. \end {aligned} $$