What is the integration of of log (1+Tan x)?

The actual question should be [0,/4] log ( 1 + tan x ) dx. say, I= [0,/4] log ( 1 + tan x ) dx.(1) now , [a,0]f(x)dx= [a,0]f(a-x)dx therefore, I = [0,/4]log[1+tan(/4-x)]dx = [0,/4]log{1+(1-tanx)/(1+tanx)}dx = [0,/4]{ log2-log(1+tanx)}dx.(2) /4 now, (1)+(2)= 2I= [0,/4](log2)dx= (log2