DUIS: ∫ log(1+a²x²)/(1+b²x²) dx

Let I(a) = ∫₀^∞ log(1 + a²x²)/(1 + b²x²) dx, with a, b > 0. Differentiate with respect to a:

dI/da = ∫₀^∞ 2a x² / [(1 + a²x²)(1 + b²x²)] dx

Partial fractions. Since 1/(1 + a²x²) − 1/(1 + b²x²) = (b² − a²)x² / [(1 + a²x²)(1 + b²x²)],

x² / [(1 + a²x²)(1 + b²x²)] = [1/(b² − a²)] [1/(1 + a²x²) − 1/(1 + b²x²)].

Using ∫₀^∞ dx/(1 + k²x²) = π/(2k):

dI/da = 2a/(b²−a²) · (π/2)(1/a − 1/b)
      = 2a/(b²−a²) · π(b−a)/(2ab)
      = π / (b(a+b))

Integrate with respect to a:

I(a) = (π/b) log(a + b) + C

When a = 0 the integrand is log 1 = 0, so I(0) = 0: C = −(π/b) log b.

Answer: ∫₀^∞ log(1 + a²x²)/(1 + b²x²) dx = (π/b) log((a + b)/b) = (π/b) log(1 + a/b).