Convergence of $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ I want to show that $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ is convergent.
I tried to use the limit convergence test by comparing with $\frac{1}{x^2}$ but that doesn't work. Any hints on how to proceed with this?

You can set $x^2=sect$ and perform this substitution. When you do that, you in up with an integral of the form$\int\frac{dt}{\sqrt{cost}}$. The key is in the upper and lower bound. For both $x=1$ and $x=2$ you can conclude that on the corresponding $t$ interval, the function is continuous. in other words, it is no longer an improper integral, the integral has an answer (though through elementary techniques we can' figure that out) thus convergent.
