${\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}x\tan x\tan \left(\frac{\pi }{4}-x\right)}dx=$

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}x\tan x\tan \left(\frac{\pi }{4}-x\right)dx,x\to \frac{\pi }{4}-x}$

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\frac{\pi }{4}-x\right)\tan \left(\frac{\pi }{4}-x\right)\tan xdx}$

$2I=\frac{\pi }{4}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\tan x\tan \left(\frac{\pi }{4}-x\right)dx}$ ,

$I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\tan x\left(\frac{1-\tan x}{1+\tan x}\right)dx,\tan x=t}$

$I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{t\left(1-t\right)}{\left(1+t\right)\left(1+t^2\right)}dt}$

   $=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{1}{t^2+1}-\frac{1}{t+1}\right)dt}$

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

 $=\frac{\pi }{8}\left(\frac{\pi }{4}-\ln 2\right)$