Evaluate : `int_0^(pi/4)log(1+tanx)dxdot`

Explanation

`"Let "I= int_(0)^(pi//4) log (1+tan x)dx " " ...(1)`
`=int_(0)^(pi//4) log{1+tan ((pi)/(4) -x)}dx`
`=int_(0)^(pi//4) log{1+(tan.(pi)/(4)-tanx)/(1+tan.(pi)/(4)tanx)}dx`
`=int_(0)^(pi//4)log{1+(1-tanx)/(1+tanx)}dx`
`=int_(0)^(pi//4)log((2)/(1+tanx))dx`
`=int_(0)^(pi//4) log 2. dx-int_(0)^(pi//4)log (1+tanx)dx`
`rArr I=log 2[x]_(0)^(pi//4) -I`
`rArr 2I=(pi)/(2) log 2rArr I=(pi)/(8) log 2`