Find the integral of $\frac{x^5+x^2+4x+\sin(x)}{64+x^6} dx$ from $-2$ to $2$ Find $$\int\limits_{-2}^{2}\frac{x^5+x^2+4x+\sin(x)}{64+x^6}dx$$
I understand this questions is trying to make the point about the integrals of symmetric functions. 
So I separated all of these to get $\frac{x^5}{64+x^6}+\frac{x^2}{64+x^6}+\frac{4x}{64+x^6}+\frac{\sin(x)}{64+x^6}$
thinking it would be better to look at them individually.
So the first part, $\frac{x^5}{64+x^6}$ is easy to integrate.
But I'm not sure how to go about the rest. 
I know that the integral of an odd function is zero because the limits of integration are symmetrical about the origin. But I'm not sure how to directly use this in the question, considering only the terms in the numerator are odd.
Please help!

$$\int^2_{-2}\dfrac{x^5+x^2+4 x+\sin(x)}{64+x^6} dx=\int^2_{-2}\dfrac{x^5}{64+x^6} dx+\int^2_{-2}\dfrac{x^2}{64+x^6} dx+\\ \int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\
\left.\dfrac{1}{6}\ln|64+x^6|\right|^2_{-2}+\int^2_{-2}\dfrac{x^2}{1+\left(\frac{x^3}{8}\right)^2} dx+\int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\
0+\left.\dfrac{1}{24}\arctan\left(\frac{x^3}{8}\right)\right|^2_{-2}+0+0\text{ (by properties of odd functions)}=\boxed{\dfrac{\pi}{48}}$$
