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!
 A: $$\int^2_{-2}\dfrac{x^5+x^2+4 x+\sin(x)}{64+x^6} dx=\int^2_{-2}\dfrac{x^5}{64+x^6} dx+\int^2_{-2}\dfrac{x^2}{64+x^6} dx+\\ \int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\
\left.\dfrac{1}{6}\ln|64+x^6|\right|^2_{-2}+\int^2_{-2}\dfrac{x^2}{1+\left(\frac{x^3}{8}\right)^2} dx+\int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\
0+\left.\dfrac{1}{24}\arctan\left(\frac{x^3}{8}\right)\right|^2_{-2}+0+0\text{ (by properties of odd functions)}=\boxed{\dfrac{\pi}{48}}$$
A: We have
$$\frac{x^5+x^2+4x+\sin(x)}{64+x^6}=\underbrace{\frac{x^5+4x+\sin(x)}{64+x^6}}_{=f(x)}+\underbrace{\frac{x^2}{64+x^6}}_{=g(x)}=f(x)+g(x)$$
Now $f(-x)=-f(x)$, so $f$ is an odd function. Also $g(x)=g(-x)$, so $g$ is an even function. That is
$$\int_{-2}^2 f(x)=0$$
and 
$$\int_{-2}^2 g(x)= 2\int_0^2 g(x) dx$$
$g$ is easy to integrate: a primitive is given by $\frac{1}{24}\arctan(x^3/8)$.
So the last integral (without the factor $2$) equals
$$\left.\frac{1}{24}\arctan(x^3/8)\right|_0^2=\frac{\arctan(1)}{24}=\frac{\pi}{96}$$
That is, your result is
$$\frac{\pi}{48}$$
A: Hint: The function $$\frac{1}{64 + x^6}$$ is even, and the product of an even function and an odd one is odd. This reduces the problem to considering only one of the terms.
