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 $a$?

$$
\begin{align}
I & = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x^a)}\\
\frac{dI}{da} & = -\int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}
\end{align}
$$
Let $\displaystyle J = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}$
$$
\begin{align}
J & = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}\\
& \stackrel{x=1/y}{=} \int_{0}^{\infty} \frac{1/y^a \log(1/y) d(1/y)}{(1+(1/y)^2)(1+(1/y)^a)^2}\\
& = \int_{\infty}^{0} \frac{y^a \log(y) dy}{(1+y^2)(1+y^a)^2}\\
& = -J
\end{align}
$$
Hence, $\frac{dI}{da}=0$. Hence, $I$ is independent of $a$.
