Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$? Let $a$ be a non-zero real number. Is it true that the value of $$\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$$  is independent on $a$?

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.
