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 $\int_0^{\infty} \frac{(x-n)^{p}}{(1+(x-n)^2)^3} dx$?
First, remove the square by letting $u=x^2$:
$$I=\frac12\int_0^{+\infty}\frac{u^{\frac{p-1}2}}{(1+u)^3}\ du$$
As Zaid Alyafeai notes, this reduces to the Euler integral of the first kind, so we have
$$I=\frac{\Gamma\left(\frac{1+p}2\right)\Gamma\left(\frac{5-p}2\right)}{2\Gamma(3)}=\frac{\pi(3-p)}{8\sin\left(\frac\pi2(1+p)\right)}$$
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.
