Value of a certain integral How to find the value of the integral
$$
\int_o^{\infty} \! \frac{x^8}{1+x^4+x^6+x^{10}} \, \mathrm{d}x
$$
given to be $\frac{1}{12}(3\sqrt{2}-1) \pi$ by WolframAlpha and, in general, is there a procedure to find the value of the definite integral of a rational function of the form $\dfrac{x^l}{p(x)}$ where $deg(p) > l > 1$ from $0$ to $\infty$ ?
 A: You can type it in here to find the answer to the problem, 
http://www.wolframalpha.com/input/?i=int_0%5Einf+%5Cfrac%7Bx%5E8%7D%7B1%2Bx%5E4%2Bx%5E6%2Bx%5E10%7Ddx
In general, these types of integral problems are generally tackled with "the calculus of residues". Marsden - Basic Complex Analysis - Chapter 4 - Page 296, gives a lovely table for these kinds of problems. For example, if $\deg(Q(x))\ge 2+\deg(P(x))$ then, 
$$\int_{-\infty}^\infty \frac{P(x)}{Q(x)} dx = 2\pi i \cdot \sum\left(\text{residues in $H^+$}\right)+\pi i \cdot\sum\left(\text{residues on $\mathbb{R}$}\right).$$ I don't know of any general methods that don't use residue theory. 
A: For the given integral, the denominator can be factored so you get 
$1+x^4+x^6+x^{10}=1+x^4+x^6(1+x^4)=(1+x^4)(1+x^6)$. 
Then, using partial fractions you can split the denominator 
$\dfrac{x^8}{(1+x^4)(1+x^6)}=\dfrac{x^2+1}{2(1+x^4)}+\dfrac{x^4-x^2-1}{2(1+x^6)}$
Which gives you two solvable (although challenging) integrals. In general, this would be my approach to solving any integral of the given form, but if it failed I would probably try using a series to approximate the functions and integrate term by term.
A: You can apply the residue theorem; in this case, the integrand is even and therefore may be extended to the entire real line.  You may show, then that the integral over the real line is equal to
$$\oint_C dz \frac{z^8}{z^{10}+z^6+z^4+1} $$,
where $C$ is a semicircle of radius $R \to \infty$ in the upper half plane, which, in turn, is equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$.  The poles inside $C$ turn out to be
$$z_1=e^{i \pi/6}$$
$$z_2=e^{i \pi/4}$$
$$z_3=e^{i \pi/2}$$
$$z_4=e^{i 3 \pi/4}$$
$$z_5=e^{i 5 \pi/6}$$
The original integral is therefore
$$i \pi \sum_{k=1}^5 \frac{z_k^5}{10 z_k^6+6 z_k^2+4 } $$
