With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that
[tex]I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx[/tex]
Replace [tex]x \to x^{\frac1\phi} = x^{\phi-1}[/tex] :
[tex]I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx[/tex]
Split the integral at x = 1. For the integral over [1, ∞), substitute [tex]x \to \frac1x[/tex] :
[tex]\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx[/tex]
The integrals involving tan⁻¹ disappear, and we're left with
[tex]I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}[/tex]
