${\displaystyle \int \tan x\tan 2xdx=}$

$={\displaystyle \int \tan x\frac{2\tan x}{1-{\tan }^{2}x}dx}$

$=2{\displaystyle \int \frac{{\tan }^{2}x}{1-{\tan }^{2}x}dx}$

$=-2{\displaystyle \int \frac{1-{\tan }^{2}x-1}{1-{\tan }^{2}x}dx}$

$=-2{\displaystyle \int 1.dx+2{\displaystyle \int \frac{1}{1-{\tan }^{2}x}dx}}$

$=-2x+2{\displaystyle \int \frac{{\cos }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}dx}$

$=-2x+{\displaystyle \int \frac{1+\cos 2x}{\cos 2x}dx}$

$=-2x+{\displaystyle \int \sec 2xdx+x}$

$=-x+\frac{1}{2}\log \left|\sec 2x-\tan 2x\right|+C$

$=-x+\frac{1}{2}\log \left|\frac{1-\sin 2x}{\cos 2x}\right|+C$

$=-x+\frac{1}{2}\log \left|\frac{\cos x-\sin x}{\cos x+\sin x}\right|+C$

$=-x+\frac{1}{2}\log \left|\tan \left(\frac{\pi }{4}-x\right)\right|+C$