Sorry to say but you're not even close to correct, and if you wrote the above then I think you need to go back to your general basics on integration:
$\displaystyle \int\frac{x^2-1}{x^4}dx=\int\left(\frac{1}{x^2}-\frac{1}{x^4}\right)dx=-\frac{1}{x}+\frac{1}{3x^3}$.
Now evaluate the above in the integral's limits and make the upper limit tend to $\displaystyle \infty$ ...the answer must be $\displaystyle \frac{2}{3}$
Tonio