<s>[INST] Solve${{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$ Please help me to evaluate the integral:
$\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$ 
Thanks.
 [/INST] Notice that:
$$x^8 = (x^6 + 1)x^2 - x^2$$
So that your integrand takes the form:
$$\int\frac{x^2}{1+x^6}dx - \int\frac{x^2}{(1+x^6)^2}dx$$
Now substitute $u = x^3, du = 3 x^2 dx$:
$$\int\frac{1}{3(1+u^2)}du - \int\frac{1}{3(1+u^2)^2}du$$
The first integral is a multiple of $\arctan(u)$, and the second can be solved dividing again into two parts:
$$\int\frac{1}{3(1+u^2)^2}du = \int\frac{1}{3(1+u^2)}du - \int\frac{u^2}{3(1+u^2)^2}du$$
I think you get the idea..
</s>