Compute the integral using the residue theorem: $\int_{-\infty}^\infty \frac{x^2}{x^6 + 1}dx$. Compute the integral using the residue theorem: $\int_{-\infty}^\infty \frac{x^2}{x^6 + 1}dx$.
If we let $\gamma_R$ be the line from $-R$ to $R$, and $\gamma_C$ be the upper half circle, and integrate ccw, we have
$$\int_{\gamma_C \cup \gamma_R} \frac{x^2}{x^6+1}dx = \int_{-R}^R \frac{x^2}{x^6+1}dx + \int_{\gamma_C}\frac{x^2}{x^6 + 1}dx.$$
Now, I need to calculate the residues of the LHS. But it seems really difficult! I see that the poles in our contour are at $x_1 = e^{\pi/6}$, $x_2 = e^{3\pi/6}$, $x_3 = e^{5\pi/6}$. Now, here is where I am getting lost:
We have $\text{res}_{x_1} = \lim_{x\to e^{\pi/6}} (x-e^{\pi/6})\frac{x^2}{x^6 + 1}.$ But if I try to split up the $x^6 + 1$ in the bottom of the fraction here, I get a bunch of terms $(x-e^{3\pi/6})(e-e^{5\pi/6})...$ etc., and that seems very messy to calculate for each pole! Is there an easier/cleaner way to do this?
 A: If you have $f(z)/g(z)$ with a simple pole in $a\in\mathbb{C}$, the residue in $a$ is simply $$\frac{f(a)}{g'(a)}.$$
Proof: since $g(a)=0$ the residue is $$\lim_{z\to a} \frac{f(z)}{g(z)}(z-a)=\lim_{z\to a}f(z)\frac{z-a}{g(z)-g(a)}$$
So in your case the residue is $\frac{a^2}{6a^5}=\frac{1}{6}a^{-4}$ and now substitute the correct values of $a$.
A: A simple alternative to the residue theorem is given by considering that:
$$ \int_{-\infty}^{+\infty}\frac{x^2}{1+x^6}\,dx = 2\int_{0}^{+\infty}\frac{x^2}{1+x^6}\,dx = 4\int_{0}^{1}\frac{x^2}{1+x^6}\,dx\tag{1}$$
and:
$$\begin{eqnarray*} \int_{0}^{1}\frac{x^2}{1+x^6}\,dx &=& \frac{1}{3}-\frac{1}{9}+\frac{1}{15}-\ldots\\ &=& \frac{1}{3}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\ldots\right)\\&=&\frac{\arctan(1)}{3}=\frac{\pi}{12},\tag{2}\end{eqnarray*}$$
hence the original integral equals $\large\color{red}{\frac{\pi}{3}}.$
A: The substitution $u=x^3$ solves the problem immediately, the solution is longer than it needs as I want to make the substitution in a proper integral, but it boils down to $\int_{-\infty}^{+\infty}\frac{x^2}{1+x^6}\,dx =\frac{1}{3} \int_{-\infty}^{+\infty}\frac{du}{1+u^2}$  :)
$$\int_{-\infty}^{+\infty}\frac{x^2}{1+x^6}\,dx = \lim_{R_1,R_2} \int_{-R_2}^{R_1}\frac{x^2}{1+x^6}\,dx = \frac{1}{3} \lim_{R_1,R_2} \int_{-R_2^3}^{R_1^3}\frac{du}{1+u^2}\\=\frac{1}{3} \lim_{R_1,R_2} (\arctan(R_2^3)+\arctan(R_1^3))=\frac{1}{3}(\frac{\pi}{2}+\frac{\pi}{2})=\frac{\pi}{3}$$
A: One other idea to make your life easier.  By using a different contour, you can simplify the calculation.  For example, consider
$$\oint_C dz \frac{z^2}{1+z^6} $$
where $C$ is now a wedge of angle $\pi/3$ in the 1st quadrant of radius $R$.  Thus, we may write the contour integral as
$$\int_0^R dx \frac{x^2}{1+x^6} + i R \int_0^{\pi/3} d\theta\, e^{i \theta} \frac{R^2 e^{i 2 \theta}}{1+R^6 e^{i 6 \theta}} + e^{i \pi/3} \int_R^0 dt \frac{e^{i 2 \pi/3} t^2}{1+t^6}$$
Note why I chose the contour to have an angle of $\pi/3$: so that the denominator of the integrand is unchanged over the different parts of the contour.
You can easily show that the second integral vanishes as $R \to \infty$.  
Now, we may invoke the residue theorem.  Note the benefit to this contour: the only pole contained in this contour is at $z=e^{i \pi/6}$. Thus, we may write
$$\int_0^{\infty} dx \frac{x^2}{1+x^6} - \int_{\infty}^0 dt \frac{t^2}{1+t^6}= i 2 \pi \frac{e^{i 2 \pi/6}}{6 e^{i 5 \pi/6}} $$
or
$$2 \int_0^{\infty} dx \frac{x^2}{1+x^6} = \int_{-\infty}^{\infty} dx \frac{x^2}{1+x^6}= \frac{\pi}{3} $$
