Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$?

Let $\mathcal{I}(a)$ denote the integral. Then $$ \begin{eqnarray} \mathcal{I}(a) &=& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_1^\infty \frac{\mathrm{d} y}{(1+y^2)(1+y^a)} \\ &\stackrel{y=1/x}{=}& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_0^1 \frac{x^a \mathrm{d} x}{(1+x^2)(1+x^a)} \\ &=& \int_0^1 \frac{1+x^a}{(1+x^2)(1+x^a)} \mathrm{d} x = \int_0^1 \frac{1}{1+x^2} \mathrm{d} x = \frac{\pi}{4} \end{eqnarray} $$

Thus $\mathcal{I}(a) = \frac{\pi}{4}$ for all $a$. I do not see a need to require $a$ to be non-zero.

With a change of variable $$ \int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)}\overset{x\to1/x}{=}\int_0^\infty\frac{x^a\mathrm{d}x}{(1+x^2)(1+x^a)} $$ Adding and dividing by two yields $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)} &=\frac{1}{2}\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\ &=\frac{\pi}{4} \end{align} $$

$\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$

Substitution:

$\displaystyle x=\tan\theta$

$\displaystyle dx=\sec^2\theta d\theta$

$\displaystyle I=\int_0^{\pi/2}\frac{d\theta}{1+\tan^a \theta}$

$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a \theta d\theta}{\sin^a \theta + \cos^a \theta}$

$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a(\pi/2-\theta) d\theta}{\sin^{a}(\pi/2-\theta) + \cos^a (\pi/2-\theta)}$

$\displaystyle I=\int_0^{\pi/2}\frac{\sin^a\theta d\theta}{\sin^a\theta + \cos^a \theta}$

Therefore,

$\displaystyle 2I=\int_0^{\pi/2}d\theta$

$\displaystyle I=\pi/4$