∫ (1-tan x)/(1+tan x)dx =

$I=\int\frac{1-tanx}{1+tanx}dx$

$=\int\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}dx=\int\frac{cosx-sinx}{cosx+sinx}dx$

Let $cosx+sinx=t$

$⇒dt=(-sinx+cosx)dx$

$⇒I=\int\frac{dt}{t}=\log|t|+C$

$=\log|cosx+sinx|+C$