The integral ∫₁⁄₂₀ ln(1+2x)/(1+4x²)dx equals : π/4ln2, π/8ln2, π/12ln2, π/32ln2

I=∫₁⁄₂₀log(1+2x)/(1+4x²)dx
Let 2x=tanθ
2dx=sec²θdθ
I=∫π/₄₀(1/2)log(1+tanθ)dθ
I=(1/2)∫π/₄₀log(1+tan(π/4−θ))dθ=(1/2)∫π/₄₀log(2/(1+tanθ))dθ=(1/2)∫π/₄₀(log2−log(1+tanθ))dθ
(3/2)I=(1/2)∫π/₄₀log2dθ
I=π/12log2