The integral is $\int_{1}^{\infty}\sqrt{\frac{x+2}{x^4-1}}dx$.
My guess is that it does converge, since we are dividing a 1st order polynomial by a 4th order one. So it is kind of like the integral of $x^{-3/2}$ which converges with these limits. Though my intuition here might be a bit off. Anyway, I have no idea how to show this (I have tried upper and lower bounds on the polynomials, setting $x+2 \leq 3x$, etc. but it gets me nowhere).
Maybe this helps: $$\begin{split} \int_1^\infty\sqrt{\frac{x+2}{x^4-1}} &\leq\int_1^\infty\sqrt{\frac{2x+2}{x^4-1}}dx \\&=\sqrt{2}\int_1^\infty\sqrt{\frac{x+1}{x^4-1}}dx \\&=\sqrt{2}\int_1^\infty\sqrt{\frac{1}{(x-1)(x^2+1)}}dx \\&\leq\sqrt{2}\int_1^\infty\sqrt{\frac{1}{(x-1)x^2}}dx \\&=\sqrt{2}\int_1^\infty\frac{1}{x\sqrt{x-1}}dx \\&=\pi\sqrt{2} \end{split}$$ where the final integral is a standard one (you can verify it by making the substitution $u=\sqrt{x-1}$ which yields a standard $\arctan$ integral).