<s>[INST] What is $\int_0^{\infty} \frac{x^{p}}{(1+x^2)^3} dx$? We know that 
$\int_0^{\infty} \frac{1}{1+x^2} dx = \frac{\pi}{2}.$
(In fact, $F(x)= \int_0^{x} \frac{1}{1+t^2}dt = [\arctan (t)]_0^{x}$, and  so $\lim_{x\to \infty} F(x) = \frac{\pi}{2}.$)
Let $p \in \{2,3,4\}, n\in \mathbb N$

How should I evaluate $\int_0^{\infty} \frac{x^{p}}{(1+x^2)^3} dx$? What is $\int_0^{\infty} \frac{(x-n)^{p}}{(1+(x-n)^2)^3} dx$?

 [/INST] In the integral with $p=3$, the subsitution $y= x^2$ works out well. For $p=2$ and $p=4$ you can use the following partial fractions expansions
$$\frac{x^2}{(1+x^2)^3} = \frac{1}{(1+x^2)^2}-\frac{1}{(1+x^2)^3} $$
$$\frac{x^4}{(1+x^2)^3} = \frac{1}{1+x^2}-\frac{2}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}$$
and the problem is reduced to find a primitive for
$$ \frac{1}{(1+x^2)^2} \quad \text{and} \quad \frac{1}{(1+x^2)^3}, $$
which can be easily done integrating
$$\int 1 \cdot \frac{1}{1+x^2}dx $$
by parts.
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