∫0π/4 log⁡1+tan2⁡θ+2tan⁡θdθ=

Let I=∫0π/4 log⁡(1+tan⁡θ)dθ=∫0π/4 log⁡(1+tan⁡(π/4−θ))dθ=∫0π/4 log⁡1+1−tan⁡θ1+tan⁡θdθ=∫0π/4 [log⁡2−log⁡(1+tan⁡θ)]dθ⇒ 2I=π/4log⁡2.Required integral =2I=π4log⁡2