Question

`\rm \int_(0)^ \infty \frac{ \sqrt[ \scriptsize\phi]{x} \tan^(- 1) (x)}{(1 + {x}^( \phi) {)}^(2) } {}^{} {} \: dx\\`​

Answer 1

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

`I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^(-1)(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty (x^(\phi-1) \tan^(-1)(x))/(x (1+x^\phi)^2) \, dx`

Replace
`x \to x^(\frac1\phi) = x^(\phi-1)` :

`I = \displaystyle \frac1\phi \int_0^\infty (\tan^(-1)(x^(\phi-1)))/((1+x)^2) \, dx`

Split the integral at x = 1. For the integral over [1, ∞), substitute
`x \to \frac1x` :

`\displaystyle \int_1^\infty (\tan^(-1)(x^(\phi-1)))/((1+x)^2) \, dx = \int_0^1 (\tan^(-1)(x^(1-\phi)))/(\left(1+\frac1x\right)^2) (dx)/(x^2) = \int_0^1 (\pi2 - \tan^(-1)(x^(\phi-1)))/((1+x)^2) \, dx`

The integrals involving tan⁻¹ disappear, and we're left with

`I = \displaystyle \frac\pi{2\phi} \int_0^1 (dx)/((1+x)^2) = \boxed{\frac\pi{4\phi}}`