<s>[INST] Evaluating $\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$ 
How to evaluate $$\int_{0}^{\infty} \frac{x^8 -1}{x^{10}  + 1} dx$$


Attempt:
$$f(x) = \frac{x^8-1}{x^{10}+1}$$
$$I = \int_{0}^{\infty} \frac{x^8 -1}{x^{10}  + 1} dx$$
Substituting $t = \frac{1}{x}$,
$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{x^{10}}  + 1} x^2dx$$
$$I = \int_{0}^{\infty} \frac{x^{4} - x^{12}}{x^{10}  + 1} dx$$
Adding the two versions of $I$,
$$2I =  \int_{0}^{\infty} \frac{x^8 - x^{12} -1 + x^4}{x^{10}  + 1} dx = \int_{0}^{\infty} \frac{(x^8-1)(1 - x^{4}) }{x^{10}  + 1} dx$$

However, that does not help much.
Putting it in WolframAlpha gives $I = 0$
I am pretty sure there is an elegant solution which I am not able to find it.
 [/INST] As @DinosaurEgg and @StefanLafon note, in fact $I=\int_0^\infty\frac{1-x^8}{x^{10}+1}dx=-I$ after the substitution, so $I$ is $0$ if it converges. Indeed, the integrand is $O(1)$ for small $x$ and $O(1/x^2)$ for large $x$.
</s>