Compute $\int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\,dx$ 
Compute the definite integral
$$
\int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\,dx
$$
where $a,b\in\mathbb{R}$.

My Attempt:
Let $x=\frac{1}{t}$ so that $dx=-\frac{1}{t^2}\,dt$. Substituting into the integral and changing the limits of integration gives
$$
\begin{align}
\int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\,dx&=\int_\infty^0\frac{t^b-t^a}{(t^a+1)(t^b+1)}\cdot\frac{-1}{t^2}\,dt\\
&=-\int_0^\infty\frac{t^a-t^b}{(1+t^a)(1+t^b)}\cdot\frac{1}{t^2}\,dt\\
&=-\int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\cdot\frac{1}{x^2}\,dx
\end{align}
$$
I'm not sure how to compute the integral from here.
 A: I must say that I am embarassed to give an answer ignoring what is your knowledge in the area of special functions. So, please, forgive me is this is out of your scope.  
The antiderivative $$I=\int \frac{dx}{1+x^a}=x \, _2F_1\left(1,\frac{1}{a};1+\frac{1}{a};-x^a\right)$$where appears the hypergeometric function. Concerning the integral $$I=\int_0^\infty \frac{dx}{1+x^a}= \frac{\pi }{a}\, \csc \left(\frac{\pi }{a}\right)$$  provided $\Re(a)>1$.
A: If you use partial fractions, you will see that the integrand is $$\frac{1}{1+x^b} - \frac{1}{1+x^a}$$ Neither summand can be indefinitely integrated in elementary terms, but the residue theorem is your friend.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{0}^{\infty}{x^{a} - x^{b} \over \pars{1 + x^{a}}\pars{1 + x^{b}}}\,\dd x
=\int_{0}^{\infty}{\dd x \over 1 + x^{b}} - \int_{0}^{\infty}{\dd x \over 1 + x^{a}}}$

Let's consider $\ds{\int_{0}^{\infty}{\dd x \over 1 + x^{\mu}}}$ with
$\Re\pars{\mu} > 1$. With the change of variables $\ds{t \equiv {1 \over 1 + x^{\mu}}}$
$\iff$ $\ds{x = \pars{1 - t \over t}^{1/\mu}}$
\begin{align}
\color{#00f}{\large\int_{0}^{\infty}{\dd x \over 1 + x^{\mu}}}&=\int_{1}^{0}
t\,{1 \over \mu}\,\pars{1 - t \over t}^{1/\mu - 1}\,\pars{-\,{\dd t \over t^{2}}}
={1 \over \mu}\int_{0}^{1}t^{-1/\mu}\pars{1 - t}^{1/\mu - 1}\,\dd t
\\[3mm]&={1 \over \mu}\,{\rm B}\pars{-\,{1 \over \mu} + 1,{1 \over \mu}}
={1 \over \mu}\,
{\Gamma\pars{-1/\mu + 1}\Gamma\pars{1/\mu}
\over \Gamma\pars{\bracks{-1/\mu + 1} + 1/\mu}}
={1 \over \mu}\,{\pi \over \sin\pars{\pi\,\bracks{1/\mu}}}
\\[3mm]&=\color{#00f}{\large{\pi \over \mu}\,\csc\pars{\pi \over \mu}}
\end{align}

Then,
$$\!\!\!\color{#00f}{\large%
\int_{0}^{\infty}\!\!\!{x^{a} - x^{b} \over \pars{1 + x^{a}}\pars{1 + x^{b}}}\,\dd x
=
{\pi \over b}\,\csc\pars{\pi \over b} - {\pi \over a}\,\csc\pars{\pi \over a}}
\,,\qquad\Re\pars{a} > 1\,,\ \Re\pars{b} > 1
$$
