<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] with $x = e^u$ and $dx = e^u du,$ I got
$$ \int_{- \infty}^\infty \frac{e^{4u} - e^{-4u}}{e^{5u} + e^{-5u}} du =  \int_{- \infty}^\infty \; \frac{ \sinh 4u \; }{ \cosh 5u \;} du$$
which is an odd integrand over symmetric endpoints
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