>>627
x = tanθ と置いて
∫[0→1]log(1+x) /(x^2+1)dx
= ∫[0→π/4] log(1+tanθ) / (tanθ^2+1) d{tanθ}
= ∫[0→π/4] log(1+tanθ) dθ
= (1/2) * { ∫[0→π/4] log(1+tanθ) dθ + ∫[0→π/4] log(1+tan(π/4-θ)) dθ }
= (1/2) * ∫[0→π/4] log(2)
= π/8 * log(2)

途中で関係式
・dx = d{tanθ} = 1/cosθ^2 * dθ = (1 + tanθ^2 ) dθ
・1+tan(π/4-θ) = 1 + (tan(π/4) - tanθ)/(1 + tan(π/4)tanθ) = 2/(1+tanθ)
を使った.