The value of ∫018log1+x 11+x2dx is

Let I=8∫01 log⁡(1+x)1+x2dx=8∫0π/4 log⁡(1+tan⁡θ)sec2⁡θsec2⁡θdθ(x=tan⁡θ)So I=8∫0π log⁡(1+tan⁡θ)dθ=8∫0π/4 log⁡1+tan⁡π4−θdθ∫0a f(x)dx=∫0a f(a−x)dx=8∫0π/4 log⁡1+1−tan⁡θ1+tan⁡θdθ=8∫0π/4 log⁡21+tan⁡θdθ=(8log⁡2)π4−I⇒ 2I=(8log⁡2)π4⇒I=πlog⁡2