<s>[INST] Test of convergence of $\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$ I am having some trouble with this problem and don't know if I am doing it right: 

$$\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$$

so the steps I have taken so far are, I split it into
$$\int_0^\infty \dfrac{x^6+6}{x^8+8} + \int_{-\infty}^0 \dfrac{x^6+6}{x^8+8}$$
for the second integral I made a change of variable $x = -u$
so it would look the same as the first integral and then use $$ \lim_{x\to \infty} \dfrac{f(x)}{g(x)}$$ where  $g(x) = \dfrac1{x^2}$.
Just wondering if my approach is correct.
 [/INST] Your approach is fine, as long as you compare with $1/x^2$ when $|x|\gt1$ and then use the continuity of $\frac{x^6+6}{x^8+8}$ for $|x|\le1$. We can illustrate this with a slightly different comparison function.
First, note that
$$
\lim_{|x|\to\infty}\frac{(x^6+6)(x^2+1)}{x^8+8}=1\tag{1}
$$
Thus, $(1)$ says that there is an $m$ so that if $|x|\ge m$, then $\frac{(x^6+6)(x^2+1)}{x^8+8}\le2$. Since $\frac{(x^6+6)(x^2+1)}{x^8+8}$ is continuous on the compact set $[-m,m]$, it is bounded there. Therefore, there is an $M$ so that
$$
\frac{(x^6+6)(x^2+1)}{x^8+8}\le M\tag{2}
$$
Therefore,
$$
\frac{x^6+6}{x^8+8}\le\frac{M}{1+x^2}\tag{3}
$$
Now, simply use
$$
\int_{-\infty}^\infty\frac{\mathrm{d}x}{1+x^2}=\pi\tag{4}
$$
to show that
$$
\int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x\tag{5}
$$
converges by comparison.

Although this answer uses contour integration, it does give the value for
$$
\int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x=\frac{\pi}{8}\csc\left(\frac{\pi}{8}\right)\left(2^{5/8}+3\cdot2^{-5/8}\right)\tag{6}
$$
</s>