### On the closed form of the generalized square-sum arctan integral

I was recently thinking about the closed-form expression to the $n$th order of the arctan integral in the case of the denominator of the integrated being a sum of squares. In context, the general sum of squares arctan integral is $$\int \dfrac{\mathrm{d}x}{x^2 + a^2} = \dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right) + C$$ $$\int_{0}^{\infty} \dfrac{\mathrm{d}x}{x^2 + a^2} = \dfrac{\pi}{2a}$$ A proof of this employs a basic substitution (factorization combined with $u = \frac{x}{a}$) and the noting of the arctan integral (a result that can be derived via implicit differentiate, with an example given on a previous post). Here, we shall consider the following integral, which we will evaluate using contour integration. $$\int_{\mathbb{R^{+}}} \dfrac{\mathrm{d}x}{x^n + a^n}$$ Factorizing $$\dfrac{1}{a^n} \int_{\mathbb{R^{+}}} \dfrac{\mathrm{d}x}{\left(\frac{x}{a}\right)^n + 1}$$ To simplify the denominator of our integrated, let $u = \frac{x}{a}$, with no change to our bounds, the reciprocal d