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With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+4 \pi C-C \ln 2-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{1}{2}i\right)+2 i \left[\,\text{Li}_2\left(1-e^{i\pi/4}\right) - \text{Li}_2\left(-e^{i\pi/4}\right)\right] \ln 2-\left(\tfrac{69}{8}-\tfrac{35 }{32}i\right) \zeta (3)-\tfrac{23}{192} \pi ^3+\left(\tfrac{7}{2}-\tfrac{7 }{8}i\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 }{32}i\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+e^{i\pi/4}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 }{8}i\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 }{4}i\right) \arctan\tfrac{1}{2}-\left(\tfrac{11}{2}-\tfrac{7 }{2}i\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.

With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+(4 \pi-\ln 2) C-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{1}{2}i\right)+2 i \left[\,\text{Li}_2\left(1-e^{i\pi/4}\right) - \text{Li}_2\left(-e^{i\pi/4}\right)\right] \ln 2-\left(\tfrac{69}{8}-\tfrac{35 }{32}i\right) \zeta (3)-\tfrac{23}{192} \pi ^3+\left(\tfrac{7}{2}-\tfrac{7 }{8}i\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 }{32}i\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+e^{i\pi/4}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 }{8}i\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 }{4}i\right) \arctan\tfrac{1}{2}-\left(\tfrac{11}{2}-\tfrac{7 }{2}i\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.

With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+4 \pi C-C \ln 2-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{i}{2}\right)+2 i \,\text{Li}_2\left(1-\tfrac{1+i}{\sqrt{2}}\right) \ln 2-2 i \,\text{Li}_2\left(-\tfrac{1+i}{\sqrt{2}}\right) \ln 2-\left(\tfrac{69}{8}-\tfrac{35 i}{32}\right) \zeta (3)-\tfrac{23 \pi ^3}{192}+\left(\tfrac{7}{2}-\tfrac{7 i}{8}\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 i}{32}\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+\tfrac{1+i}{\sqrt{2}}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 i}{8}\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 i}{4}\right) \arctan\left(\tfrac{1}{2}\right)-\left(\tfrac{11}{2}-\tfrac{7 i}{2}\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.

With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+4 \pi C-C \ln 2-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{1}{2}i\right)+2 i \left[\,\text{Li}_2\left(1-e^{i\pi/4}\right) - \text{Li}_2\left(-e^{i\pi/4}\right)\right] \ln 2-\left(\tfrac{69}{8}-\tfrac{35 }{32}i\right) \zeta (3)-\tfrac{23}{192} \pi ^3+\left(\tfrac{7}{2}-\tfrac{7 }{8}i\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 }{32}i\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+e^{i\pi/4}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 }{8}i\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 }{4}i\right) \arctan\tfrac{1}{2}-\left(\tfrac{11}{2}-\tfrac{7 }{2}i\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.

With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\log(1+x^2)-\log(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \, _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+4 \pi C-C \log 2-2 i \text{Li}_3\left(\tfrac{1}{2}+\tfrac{i}{2}\right)+2 i \text{Li}_2\left(1-\tfrac{1+i}{\sqrt{2}}\right) \log 2-2 i \text{Li}_2\left(-\tfrac{1+i}{\sqrt{2}}\right) \log 2-\left(\tfrac{69}{8}-\tfrac{35 i}{32}\right) \zeta (3)-\tfrac{23 \pi ^3}{192}+\left(\tfrac{7}{2}-\tfrac{7 i}{8}\right) \pi +\tfrac{1}{24} i \log ^3 2-\tfrac{7}{16} \pi \log ^2 2-\left(\tfrac{1}{12}+\tfrac{9 i}{32}\right) \pi ^2 \log 2)+\tfrac{1}{2} \pi \log \left(1+\tfrac{1+i}{\sqrt{2}}\right) \log 2+\left(\tfrac{7}{8}+\tfrac{11 i}{8}\right) \log 5-\left(\tfrac{3}{2}+\tfrac{7 i}{4}\right) \arctan\left(\tfrac{1}{2}\right)-\left(\tfrac{11}{2}-\tfrac{7 i}{2}\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$

with $C$ Catalan's constant and $\text{Li}_n$ the polylog.

With some effort, Mathematica evaluates this as

$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+4 \pi C-C \ln 2-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{i}{2}\right)+2 i \,\text{Li}_2\left(1-\tfrac{1+i}{\sqrt{2}}\right) \ln 2-2 i \,\text{Li}_2\left(-\tfrac{1+i}{\sqrt{2}}\right) \ln 2-\left(\tfrac{69}{8}-\tfrac{35 i}{32}\right) \zeta (3)-\tfrac{23 \pi ^3}{192}+\left(\tfrac{7}{2}-\tfrac{7 i}{8}\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 i}{32}\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+\tfrac{1+i}{\sqrt{2}}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 i}{8}\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 i}{4}\right) \arctan\left(\tfrac{1}{2}\right)-\left(\tfrac{11}{2}-\tfrac{7 i}{2}\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.