Integration of a rational function from +/- infinity I am trying to calculate the integral
$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$
where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

Hint: Use partial fraction to write the integral as
$$ {\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}} = \frac{A+Bx}{b^2 + (a+x)^2} + \frac{C+D x}{1+c(a-x)^2}, $$
then you can write it as
$$ \frac{A}{b^2 + (a+x)^2} + \frac{C}{1+c(a-x)^2} + \frac{Bx}{b^2 + (a+x)^2} + \frac{D x}{1+c(a-x)^2}, $$
and finally make the change of variables $x-a=t$ and $x+a=t$ and things will be easy. See here for similar techniques and read the comments under my answer.
