Prove that integral is independant of its parameter We are given intergral $\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}}$ and task is to prove that it's independent of $\alpha$.
Task was too complicated for me, so I had to stick with solution, recommended in the textbook. It goes on like that:
$\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_0^1 {\frac {dx} {(1+x^2)(1+x^\alpha)}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = J_1 + J_2$
In $J_1$ we do replacement $y = \frac 1 x$
$J_1 = \int_1^\infty \frac {dy} {-(1+y^2)(1+y^{-\alpha})}$  
Than textbook says that next step is
$J_1+J_2 = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha})  \frac {dx}{1+x^2}}$
From where it's obvious that inital statement is true. Problem is, I completely fail to understand how that summation of $J_1+J_2$ was done.
Edit
To be more explicit, I can't get this: $\int_1^\infty{\frac {dy}{-(1+y^2)(1+y^{-\alpha})}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha})  \frac {dx}{1+x^2}}$  
So I would appreciate some explanations and pointers.

There is no need to break the integral into pieces. Simply use the substitution $t=1/x$:
$$
\begin{align}
A
&=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\tag{1}\\
&=\int_0^\infty\frac{t^\alpha\,\mathrm{d}t}{(1+t^2)(1+t^\alpha)}\tag{2}
\end{align}
$$
Add $(1)$ and $(2)$ to get
$$
\begin{align}
2A
&=\int_0^\infty\frac{(1+x^\alpha)\,\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\\
&=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\
&=\frac\pi2\tag{3}
\end{align}
$$
