Why is it that $\int_{-\infty}^{\infty} \frac{x}{1+x^2} dx $ diverges, while $\int_{-\infty}^{\infty} \frac{x}{1+x^8} dx $ converges. Why does $$ \int_{-\infty}^{\infty} \frac{x}{1+x^2} dx $$ diverge, while $$ \int_{-\infty}^{\infty} \frac{x}{1+x^8} dx $$ converges?
 A: Note that, for $x\geq 1$, we have that
$$\frac{1}{2x} \leq \frac{x}{1+x^2}$$
It follows that the integral
$$\int_0^\infty \frac{x}{1+x^2} dx$$
diverges, and so does the original integral. As for the second integral, note that
$$0 \leq \frac{x}{1+x^8} \leq \frac{1}{x^7}$$
and so
$$\int_{-\infty}^\infty \frac{x}{1+x^8} dx = \int_{-\infty}^{-1} \frac{x}{1+x^8}dx + \int_{-1}^1 \frac{x}{1+x^8}dx + \int_{1}^\infty \frac{x}{1+x^8}dx$$
As we can see, the first and last terms converge by the comparison test, and the middle term is a proper integral of a continuous function and thus converges. So the whole integral converges. 
A: The short answer is that
$$\frac{x}{1+x^2} \sim \frac1x $$
when $x\to\infty$.
I assume (perhaps incorrectly) that you know that
$$\int_1^\infty \frac1x\,\mathrm d x$$
diverges.
A: A primitive for $\frac{x}{1+x^2}$ is given by $\frac{1}{2}\log(1+x^2)$, that is an unbounded function as $x\to\pm\infty$.
On the other hand $\frac{x}{1+x^8}$ is bounded by $1$ in absolute value over the interval $[-1,1]$, and by $\frac{1}{x^7}$ on the intervals $(-\infty,-1)$ and $(1,+\infty)$. It follows that $\frac{x}{1+x^2}\not\in L^1(\mathbb{R})$ but $\frac{x}{1+x^8}\in L^1(\mathbb{R})$, and since it is an odd function its integral over $\mathbb{R}$ simply equals zero.
