DUIS: ∫ tan⁻¹(ax)/(x(1+x²)) dx

Note: some copies print the upper limit as α; the intended limit is ∞ (with a finite upper limit the integral has no neat closed form).

Let I(a) = ∫₀^∞ tan⁻¹(ax)/(x(1 + x²)) dx. Differentiate with respect to a:

∂/∂a tan⁻¹(ax) = x/(1 + a²x²), so

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

Partial fractions (a ≠ 1):

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

dI/da = [1/(1−a²)] [π/2 − a²·π/(2a)]
      = [1/(1−a²)] (π/2)(1 − a)
      = π / (2(1 + a))

Integrate: I(a) = (π/2) log(1 + a) + C. At a = 0, I = 0, so C = 0.

Answer: ∫₀^∞ tan⁻¹(ax) / (x(1 + x²)) dx = (π/2) log(1 + a).