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 use the fact that $\frac{1}{\sqrt{x^4-1}} < \frac{1}{\sqrt{x-1}}$ for all $x>1$. This is because $x^4-1>x-1$ for all $x>1$.
Therefore, we have $\int_1^2 \frac{1}{\sqrt{x^4-1}} < \int_1^2 \frac{1}{\sqrt{x-1}} = 2\sqrt{x-1} |_1^2 = 2$
