${\int }_0^{\frac{\pi }{4}}x\tan x\tan \left(\frac{\pi }{4}-x\right)dx=$

$I={\int }_0^{\frac{\pi }{4}}x\tan x\tan \left(\frac{\pi }{4}-x\right)dx,x\to \frac{\pi }{4}-x$

$I={\int }_0^{\frac{\pi }{4}}\left(\frac{\pi }{4}-x\right)\tan \left(\frac{\pi }{4}-x\right)\tan xdx$

$2I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{4}}\tan x\tan \left(\frac{\pi }{4}-x\right)dx$

$I=\frac{\pi }{8}{\int }_0^{\frac{\pi }{4}}\tan x\left(\frac{1-\tan x}{1+\tan x}\right)dx,\tan x=t$

$l=\frac{\pi }{8}{\int }_0^1\frac{t(1-t)}{(1+t)\left(1+t^2\right)}dt=\frac{\pi }{8}{\int }_0^1\left(\frac{1}{t^2+1}-\frac{1}{t+1}\right)dt$

$=\frac{\pi }{8}\left({\tan }^{-1}t-{\left.\ln (t+1)\right|}_0^1=\frac{\pi }{8}\left(\frac{\pi }{4}-\ln 2\right)\right.$