${\int }_0^1 \mathrm{x}{\left({\tan }^{-1}\mathrm{x}\right)}^2\mathrm{dx}$ is equal to

$\mathrm{I}={\int }_0^1 \mathrm{x}{\left({\tan }^{-1}\mathrm{x}\right)}^2\mathrm{dx}$

On integrating by parts, we have

$\begin{array}{c}\mathrm{I}=\frac{{\mathrm{x}}^2}{2}{\left[{\left({\tan }^{-1}\mathrm{x}\right)}^2\right]}_0^1-\frac{1}{2}{\int }_0^1 {\mathrm{x}}^2\cdot 2\frac{{\tan }^{-1}\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{{\mathrm{\pi }}^2}{32}-{\int }_0^1 \frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\cdot {\tan }^{-1}\mathrm{xdx}\\ =\frac{{\mathrm{\pi }}^2}{32}-{\mathrm{I}}_1,\text{ where }{\mathrm{I}}_1={\int }_0^1 \frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}{\tan }^{-1}\mathrm{xdx}\end{array}$

$\begin{array}{l}\mathrm{now}, {\mathrm{I}}_1={\int }_0^1 \frac{{\mathrm{x}}^2+1-1}{1+{\mathrm{x}}^2}{\tan }^{-1}\mathrm{xdx}\\ ={\int }_0^1 {\tan }^{-1}\mathrm{xdx}-{\int }_0^1 \frac{1}{1+{\mathrm{x}}^2}{\tan }^{-1}\mathrm{xdx}\\ ={\mathrm{I}}_2-\frac{1}{2}{\left\{\left({\tan }^{-1}{\mathrm{x}}^2\right)\right\}}_0^1={\mathrm{I}}_2-\frac{{\mathrm{\pi }}^2}{32}\\ \mathrm{Here}, {\mathrm{I}}_2={\int }_0^1 {\tan }^{-1}\mathrm{xdx}={\left[{\mathrm{xtan}}^{-1}\mathrm{x}\right]}_0^1-{\int }_0^1 \frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\left[\log \left|1+{\mathrm{x}}^2\right|\right]}_0^1=\frac{\mathrm{\pi }}{4}-\frac{1}{2}\log 2\\ \mathrm{Thus}, {\mathrm{I}}_1=\frac{\mathrm{\pi }}{4}-\frac{1}{2}\log 2-\frac{{\mathrm{\pi }}^2}{32}\end{array}$

Therefore,

$\begin{array}{r}{\mathrm{I}}_1=\frac{{\mathrm{\pi }}^2}{32}-\frac{\mathrm{\pi }}{4}+\frac{1}{2}\log 2+\frac{{\mathrm{\pi }}^2}{32}=\frac{{\mathrm{\pi }}^2}{16}-\frac{\mathrm{\pi }}{4}+\frac{1}{2}\log 2\\ =\frac{{\mathrm{\pi }}^2-4\mathrm{\pi }}{16}+\log \sqrt{2}\end{array}$