Putting x = tan θ turns the integral into

-2 integral of log cos θ dθ from 0 to π/2

2 integral of log sin θ dθ from 0 to π

integral of log cos θ dθ from 0 to ∞

-integral of log tan θ dθ from 0 to π/2

J (log cos) equals the integral of log sin because

of the property f(θ) → f(a-θ) with a = π/2

sin θ = cos θ on the interval

Short answer

Put x = tanθ, then I = -2∫₀^{π/2} log cosθ dθ. Using the known result ∫₀^{π/2} log cosθ dθ = −(π/2) log 2, we find I = π log 2.

Show that ∫₀^∞ log(1+x²)/(1+x²) dx = π log 2.

Let I = ∫₀^∞ log(1 + x²)/(1 + x²) dx.

Put x = tan θ, dx = sec²θ dθ, 1 + x² = sec²θ. Limits: 0 → 0, ∞ → π/2.

I = ∫₀^(π/2) log(sec²θ) dθ = −2 ∫₀^(π/2) log cos θ dθ.

Evaluate J = ∫₀^(π/2) log cos θ dθ. By the property ∫₀^a f(θ)dθ = ∫₀^a f(a − θ)dθ, J = ∫₀^(π/2) log sin θ dθ. Add:

2J = ∫₀^(π/2) log(sinθ cosθ) dθ
   = ∫₀^(π/2) [log sin 2θ − log 2] dθ
   = ∫₀^(π/2) log sin 2θ dθ − (π/2) log 2

Put 2θ = u: ∫₀^(π/2) log sin 2θ dθ = (1/2)∫₀^π log sin u du = ∫₀^(π/2) log sin u du = J. So 2J = J − (π/2) log 2, giving

J = −(π/2) log 2.

I = −2J = −2 · (−(π/2) log 2) = π log 2.

Proved: ∫₀^∞ log(1 + x²)/(1 + x²) dx = π log 2.

✓ Answer checked by Claude Opus 5.5 (high effort)