$\int \frac{2\tan \left(x\right)}{1+2{\tan }^2\left(x\right)}dx=$

$I=\int \frac{2\tan x}{1+2{\tan }^{2}x}dx=\int \frac{2\sin x}{\cos x\left(1+2\frac{{\sin }^{2}x}{{\cos }^{2}x}\right)}dx=\int \frac{2\sin x{\cos }^{2}x}{\cos x\left({\cos }^{2}x+2{\sin }^{2}x\right)}dx$

$=\int \frac{2\sin x\cos x}{\left({\cos }^{2}x+2{\sin }^{2}x\right)}dx=\int \frac{\sin 2x}{\left({\cos }^{2}x+2{\sin }^{2}x\right)}dx$

Put ${\cos }^{2}x+2{\sin }^{2}x=t$. Then $(-\sin 2x+2\sin 2x)dx=dt\Rightarrow \sin 2xdx=dt$

$\therefore I=\int \frac{1}{t}dt=\log \left|t\right|+c=\log \left|{\cos }^{2}x+2{\sin }^{2}x\right|+c=\log \left|\frac{{\cos }^{2}x}{2}+{\sin }^{2}x\right|+c\text{. }$