<s>[INST] 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?
 [/INST] Recurrence (3) in my other answer on this page also follows immediately from the same recurrence for the respective integrands! :-)
Concerning the case $n=0$: the recurrence for the $M_k$'s follows immediately from the definition of $M_k$ and the trivial identity $y^{-k}-y^{-k-1}=y^{-k-1}(y-1)$.

To make this answer independent of the previous one, note that for the integral in question we have
\begin{equation*}
    I_{n,N}=\int_0^\infty K_{n,N}(x)\,dx,
\end{equation*}
where
\begin{equation*}
    K_{n,N}(x):=\frac{x^{2n+1}}{(1+x^2)^2(2+x^2)^N}. 
\end{equation*}
Note next that $K_{n,N}=K_{n-1,N-1}-2K_{n-1,N}$ and hence
\begin{equation*}
    I_{n,N}=I_{n-1,N-1}-2I_{n-1,N} \quad \text{if $1\le n\le N$}.\tag{1}
\end{equation*}
Also,
\begin{equation*}
    2I_{0,N}=J_N:=\int_0^\infty\frac{2x\,dx}{(1+x^2)^2(2+x^2)^N}
    =\int_2^\infty\frac{y^{-N}\,dy}{(y-1)^2}.
\end{equation*}
Next,
\begin{align*}
    J_{N-1}-2J_N+J_{N+1}&=\int_2^\infty\frac{y^{-N+1}-2y^{-N}+y^{-N-1}}{(y-1)^2}\,dy \\ 
    &=\int_2^\infty y^{-N-1}\,dy=\frac1{N2^N}. 
\end{align*}
So,
\begin{equation*}
    I_{0,N+1}=2I_{0,N}-I_{0,N-1}+\frac1{N2^{N+1}}\quad\text{for $N\ge1$}, \tag{2}
\end{equation*}
with
\begin{equation*}
\text{$I_{0,0}=1/2$ and $I_{0,1}=(1-\ln2)/2$.} \tag{3}  
\end{equation*}
Formulas (1), (2), (3) provide the desired recurrence relations for $I_{n,N}$.
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