# Calculate int_0^1(log^2x log(1+x^2))/(1-x^2)dx

Calculate ${\int }_{0}^{1}\frac{{\mathrm{log}}^{2}x\mathrm{log}\left(1+{x}^{2}\right)}{1-{x}^{2}}dx$
I found $-\frac{{\pi }^{4}}{32}+2{G}^{2}+\frac{7}{4}\zeta \left(3\right)\mathrm{log}2$ where $G$ is the Catalan's constant.
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dasse9
$\begin{array}{rl}J& ={\int }_{0}^{1}\frac{{\mathrm{log}}^{2}\left(x\right)\mathrm{log}\left(1+{x}^{2}\right)}{1-{x}^{2}}dx\\ & \stackrel{\text{IBP}}{=}{\left[\left({\int }_{0}^{x}\frac{{\mathrm{ln}}^{2}t}{1-{t}^{2}}dt\right)\mathrm{ln}\left(1+{x}^{2}\right)\right]}_{0}^{1}-{\int }_{0}^{1}\frac{2x}{1+{x}^{2}}\left({\int }_{0}^{x}\frac{{\mathrm{ln}}^{2}t}{1-{t}^{2}}dt\right)dx\\ & =\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2-{\int }_{0}^{1}{\int }_{0}^{1}\frac{2{x}^{2}{\mathrm{ln}}^{2}\left(tx\right)}{\left(1+{x}^{2}\right)\left(1-{t}^{2}{x}^{2}\right)}dtdx\\ & =\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+2{\int }_{0}^{1}{\int }_{0}^{1}\left(\frac{{\mathrm{ln}}^{2}\left(tx\right)}{\left(1+{t}^{2}\right)\left(1+{x}^{2}\right)}-\frac{{\mathrm{ln}}^{2}\left(tx\right)}{\left(1+{t}^{2}\right)\left(1-{t}^{2}{x}^{2}\right)}\right)dtdx\\ & =\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+4{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}x}{\left(1+{t}^{2}\right)\left(1+{x}^{2}\right)}dtdx+4{\left({\int }_{0}^{1}\frac{\mathrm{ln}x}{1+{x}^{2}}dx\right)}^{2}-\\ & 2{\int }_{0}^{1}{\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}\left(tx\right)}{\left(1+{t}^{2}\right)\left(1-{t}^{2}{x}^{2}\right)}dtdx\\ & =\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+4×\frac{1}{16}{\pi }^{3}×\frac{1}{4}\pi +4{\text{G}}^{2}-{\int }_{0}^{1}\frac{2}{t\left(1+{t}^{2}\right)}\left({\int }_{0}^{t}\frac{{\mathrm{ln}}^{2}u}{1-{u}^{2}}du\right)dt\\ & \stackrel{\text{IBP}}{=}\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+\frac{1}{16}{\pi }^{4}+4{\text{G}}^{2}-{\left[\left(2\mathrm{ln}t-\mathrm{ln}\left(1+{t}^{2}\right)\right)\left({\int }_{0}^{t}\frac{{\mathrm{ln}}^{2}u}{1-{u}^{2}}du\right)\right]}_{0}^{1}+\\ & {\int }_{0}^{1}\frac{\left(2\mathrm{ln}t-\mathrm{ln}\left(1+{t}^{2}\right)\right){\mathrm{ln}}^{2}t}{1-{t}^{2}}dt\\ & =\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+\frac{1}{16}{\pi }^{4}+4{\text{G}}^{2}+\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+2{\int }_{0}^{1}\frac{{\mathrm{ln}}^{3}t}{1-{t}^{2}}dt-J\\ J& =\overline{)\frac{7}{4}\zeta \left(3\right)\mathrm{ln}2+2{\text{G}}^{2}-\frac{1}{32}{\pi }^{4}}\end{array}$
NB: i assume that,
$\begin{array}{rl}{\int }_{0}^{1}\frac{\mathrm{ln}t}{1+{t}^{2}}dt& =-\text{G}\\ {\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}t}{1+{t}^{2}}dt& =\frac{{\pi }^{3}}{16}\\ {\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}t}{1-{t}^{2}}dt& =\frac{7}{4}\zeta \left(3\right)\\ {\int }_{0}^{1}\frac{{\mathrm{ln}}^{3}t}{1-{t}^{2}}dt& =-\frac{{\pi }^{4}}{16}\end{array}$
Addendum: typo fixed. Thanks Sewer Keeper.