Value of $ \int _0^{+\infty }\frac{1}{\left(1+x^2\right)\left(1+x^{\alpha }\right)}dx$ while $\alpha$ is constant 
Value of $$\int _0^{+\infty }\:\frac{1}{\left(1+x^2\right)\left(1+x^{\alpha }\right)}dx$$ while  $\alpha$ is constant.

I have no idea. How to solve it?  Any suggestion?
==========update=======
Just now I read several answers, I mistakenly thought that this question is far beyond my understanding, so I accepted the answer that I did not understand and wanted to give up the question.
But, I found out:

let x = tanu

then:

$\int _0^{+\infty }\:\frac{1}{\left(1+x^2\right)\left(1+x^{\alpha }\right)}dx$
$=\int _0^{\frac{\pi }{2}}\:\frac{\sec^2u}{\left(1+\tan^2u\right)\left(1+\tan^{\alpha }u\right)}d\tan u$
$=\int _0^{\frac{\pi }{2}}\:\frac{1}{1+x^{\alpha }}dx$

It seemed a little easier. What should I do next?
 A: To expand on @EdwardH.'s comment:
Begin with$$\begin{align}\int_0^\infty f(x)dx&=\int_0^1f(x)dx+\int_1^\infty f(x)dx\\&=\int_0^1f(x)dx+\int_0^1\frac{f(1/x)}{x^2}dx\\&=\int_0^1\left(f(x)dx+\frac{f(1/x)}{x^2}\right)dx,\end{align}$$where the penultimate $=$ substitutes $x\mapsto 1/x$ in the integral on $[1,\,\infty)$. In particular, if $f(x)=g(x)/(1+x^2)$ this simplifies, since $\frac{1/x^2}{1+(1/x)^2}=\frac{1}{1+x^2}$, to$$\int_0^\infty\frac{g(x)dx}{1+x^2}=\int_0^1\frac{g(x)+g(1/x)}{1+x^2}dx.$$The choice $g(x):=\tfrac{1}{1+x^\alpha}$ satisfies$$g(x)+g(1/x)=\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}=1,$$so the requested integral is $\int_0^1\tfrac{dx}{1+x^2}=\tfrac{\pi}{4}$, regardless of $\alpha$.
A: The Mellin transform
$$
\mathcal{M}_x\left[\frac{1}{1+x^\alpha}\right](s) = \int_0^\infty x^{s-1}\frac{1}{(1+x^\alpha)} \;dx = \frac{\pi}{\alpha}\csc\left(\frac{\pi s}{\alpha}\right)
$$
we can see the other term as an instruction to sum over alternating squared terms
$$
\frac{1}{(1+x^2)} = 1 - x^2 + x^4 - x^6 + \cdots
$$
so in effect 
$$
\int_0^\infty (1 - x^2 + x^4 - x^6 + \cdots) \frac{1}{(1+x^\alpha)} \; dx = \sum_{k=0}^\infty (-1)^k \frac{\pi}{\alpha} \csc\left(\frac{\pi (2k+1)}{\alpha}\right)
$$
but the sum doesn't necessarily converge. 
A: If $\alpha=n\in\mathbb{Z}$ is integer, then this integral can be calculated using a contour integral on a complex plane:
$$ \int_0^\infty \frac{1}{(1+x^2)(1+x^n)} dx = \frac{1}{-2\pi i}\oint_\text{keyhole} \frac{\log z}{(1+z^2)(1+z^n)} dz = \\
= -\sum_{z_k} {\rm Res}\left(\frac{\log z}{(1+z^2)(1+z^n)},z_k \right)$$
where $z_k$ are all singularites of function $\frac{\log z}{(1+z^2)(1+z^n)}$.
If $\alpha=\frac{p}{q}\in \mathbb{Q}$ then by substitution $x=y^q$ we have
$$ \int_0^\infty \frac{1}{(1+x^2)(1+x^\frac{p}{q})} dx = \int_0^\infty \frac{q y^{q-1}}{(1+y^{2q})(1+y^p)} dy $$
and this integral can also be calulated by contour integration.
