For which $a\in\mathbb{R}$ this integral is convergent?
$$I=\int_0^{\infty}\frac{dx}{x^{a}+x^{-a}}$$
Note that for $a>0$
$$\frac{1}{x^{a}+x^{-a}} = \frac{x^a}{x^{2a}+1} \sim \frac{1}{x^{a}}$$
when $x\rightarrow \infty$.
So we know that $$\int_1^{\infty}\frac{dx}{x^{a}+x^{-a}}$$ is convergent only if $a>1$.
Now we need to consider $$I=\int_0^{1}\frac{dx}{x^{a}+x^{-a}}$$
Obviously:
$$\lim_{x\rightarrow 0^+}\frac{1}{x^{a}+x^{-a}}=0$$
for $a>1$, so we can conclude that $I$ is convergent whenever $a>1$.
I write this answer just to confirm that your way of working this problem is correct.
Edit
I just noted that the question asks for $a\in\mathbf R$ and not $a>0$ (as I first read). Note, however, that the integral is even, so you actually only have to consider $a>0$. The final result is that the integral is convergent if and only if $|a|>1$. This should be considered in your solution.
Comment As a curiosity, the integral can be calculated (as indicated by @Lucian), and equals $$ \frac{\pi}{2a}\sec\Bigl(\frac{\pi}{2a}\Bigr),\quad a>1. $$