Solving the inequality ${x^2+2x+2^{|a|}\over x^2-a^2} > 0$ How do I solve the inequality ${x^2+2x+2^{|a|}\over x^2-a^2} > 0$?  
My idea is that for $a\ne 1$, the numerator will always be positive. So the inequality reduces to $
{1 \over x^2-a^2 } > 0
$
My doubt is in this part. If we factorise the denominator, we get
$
{1 \over (x+a)(x-a)}
$
which, by using the wavy curve method, gives me,
$
x \epsilon (-\infty,-a) \  \ \ \cup \ \ \ (a,\infty).
$
But according to my textbook, the answer with $a\ne1$ is 
$
x \epsilon (-\infty,-|a|) \  \ \ \cup \ \ \ (|a|,\infty).
$
Why?

Your textbook is right.
For instance, take $a=-1$. Your solution would be $(-\infty,1)\cup(-1,\infty)=\mathbb R$, which is wrong.
In fact, the solution must be in terms of $|a|$, as the sign of $a$ plays no role.
