<s>[INST] integrate $\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}} dx$ So, I need to integrate $\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}} dx, n \in \mathbb{N}$.
I've tried partial fractions, but I get nowhere.
Then I tried using substitution, using the substitution $t = 1+x^2$, but I get:
$$
\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}}dx = \int \frac{x^{(2n-1)}}{(t)^{(n+3)}} \frac{1}{2x}dt =2\int int\frac{x^{(2n-2)}}{(t)^{(n+3)}}dt = 2\int\frac{(t-1)^{(n-1)}}{(t)^{(n+3)}}dt $$
And then I don't know what to do.
Should I use another approach, or am I on the right path?
Thanks in advance.
 [/INST] Denote
$$
I_n = \int \frac{x^{2n-1}}{(1+x^2)^{n+3}} dx.
$$
Using integration by parts it's easy to see that
$$
I_n = \int \frac{x^{2n-1}}{(1+x^2)^{n+3}} dx = \begin{bmatrix} u = \frac{1}{(1+x^2)^{n+3}} & dv = x^{2n-1} dx \\ du = - \frac{2(n+3)x}{(1+x^2)^{n+4}} dx & v =\frac{x^{2n}}{2n}\end{bmatrix} = 
$$
$$
= \frac{x^{2n}}{2n (1+x^2)^{n+3}} + \frac{n+3}{n} \int \frac{x^{2n+1}}{(1+x^2)^{n+4}} = \frac{x^{2n}}{2n (1+x^2)^{n+3}} + \frac{n+3}{n} I_{n+1}. 
$$
Therefore,
$$
I_{n+1} = \frac{n}{n+3} \left(I_n - \frac{x^{2n}}{2n (1+x^2)^{n+3}} \right)
$$
and one can compute $I_n$ for any $n>1$ knowing $I_1$.
</s>