If ${\int }_0^1 {\mathrm{xtan}}^{-1}\mathrm{xdx}=$$\frac{\mathrm{\pi }-k}{4}$ then k=

$$\begin{gathered} \int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx=\int \left(f\left(x\right)\int g\left(x\right)dx\right)dx \\ {\int }_0^1 x{\tan }^{-1}xdx={\int }_0^1 \left({\tan }^{-1}x\right)xdx \\ ={\tan }^{-1}x{\int }_0^1 xdx-{\int }_0^1 \left(\frac{d\left({\tan }^{-1}x\right)}{dx}{\int }_0^{1}x.dx\right)dx \\ ={\left[{\tan }^{-1}x\cdot \frac{x^2}{2}\right]}_0^1-{\int }_0^1 \frac{1}{1+x^2}dx \\ ={\left[\frac{x^2}{2}{\tan }^{-1}x\right]}_0^1-\frac{1}{2}{\int }_0^1 \frac{x^2+1-1}{x^2+1}dx \\ ={\left[\frac{x^2}{2}{\tan }^{-1}x\right]}_0^1-\frac{1}{2}\left[{\int }_0^1 dx-{\int }_0^1 \frac{dx}{x^2+1}\right] \\ ={\left[\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}x+\frac{1}{2}+\frac{1}{1}{\tan }^{-1}\frac{x}{1}\right]}_0^1 \\ =\left[\frac{(1{)}^2}{2}{\tan }^1\left(1\right)-\frac{1}{2}+\frac{1}{2}{\tan }^{-1}1\right]-\left[\frac{0^2}{2}{\tan }^{-1}0-\frac{0}{2}+\frac{1}{2}{\tan }^{-1}\left(0\right)\right] \\ =\frac{{\displaystyle \pi }}{4}-\frac{1}{2} \end{gathered}$$

$\frac{\mathrm{\pi }-2}{4}$