Show ∫ log(1+x)/(1+x²) dx = (π/8) log 2

Let I = ∫₀¹ log(1 + x)/(1 + x²) dx.

Put x = tan θ, dx = sec²θ dθ, 1 + x² = sec²θ. Limits: 0 to π/4.

I = ∫₀^(π/4) log(1 + tan θ) dθ

Use ∫₀^a f(θ)dθ = ∫₀^a f(a − θ)dθ with a = π/4. Since tan(π/4 − θ) = (1 − tan θ)/(1 + tan θ):

1 + tan(π/4 − θ) = 2/(1 + tan θ)

I = ∫₀^(π/4) log[2/(1 + tan θ)] dθ
  = ∫₀^(π/4) log 2 dθ − ∫₀^(π/4) log(1 + tan θ) dθ
  = (π/4) log 2 − I

So 2I = (π/4) log 2.

I = (π/8) log 2. Shown.