Improper integral $\int_0^\infty\frac{x^{2n+1}}{(1+x^2)^2(2+x^2)^N}dx,\ \ \ n\le N$ How can I evaluate this integral?
$$\int_0^\infty\frac{x^{2n+1}}{(1+x^2)^2(2+x^2)^N}dx,\ \ \ n\le N$$
Maybe there is a recurrence relation for the integral?

One approach is to consider the sum
$$ J = \sum_{n,m=0}^\infty s^nt^mI_{n,n+m} = \int_{x=0}^\infty F\,dx, $$
where
$$ F = \sum_{n,m=0}^\infty s^nt^m \frac{x^{2n+1}}{(1+x^2)^2(2+x^2)^{n+m}} = 
 \frac{x(2+x^2)^2}{(1+x^2)^2(2+x^2-sx^2)(2+x^2-t)}
$$
This works out as
$$
 J = \frac{2s^2\ln(1-s)}{(1+s)^2(st-2s-t)} + 
  \frac{t^2\ln(1-t/2)}{2(1-t)^2(st-2s-t)} + 
  \frac{(2s-t-3st)\ln(2)}{2(1+s)^2(1-t)^2} + 
  \frac{1}{2(1+s)(1-t)} 
$$
In principle you could expand this as a power series and extract $I_{n,n+m}$ as the coefficient of $s^nt^m$.
