show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges Show $\int_{1}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges.
I was able to show $\int_{2}^{\infty} \frac{1}{\sqrt{x^4-x}} dx$ converges, comparing it with the function $\frac{1}{x^{3/2}}$.  I have trouble showing that $\int_{1}^{2} \frac{1}{\sqrt{x^4-x}} dx$ converges due to the fact that the function is not continuous at 1.  Not sure how to do this, since I can't evaluate the integral directly.

The problem seeming to be at $x=1$, make the Taylor series
$$\frac{1}{\sqrt{x^{4} - x}} =\frac{1}{\sqrt{3} \sqrt{x-1}}-\frac{\sqrt{x-1}}{\sqrt{3}}+\frac{5 (x-1)^{3/2}}{6
   \sqrt{3}}-\frac{2 (x-1)^{5/2}}{3 \sqrt{3}}+\frac{13 (x-1)^{7/2}}{24
   \sqrt{3}}+O\left((x-1)^{9/2}\right)$$ Integrate termwise to get
$$\frac{2 \sqrt{x-1}}{\sqrt{3}}-\frac{2 (x-1)^{3/2}}{3 \sqrt{3}}+\frac{(x-1)^{5/2}}{3
   \sqrt{3}}-\frac{4 (x-1)^{7/2}}{21 \sqrt{3}}+\frac{13 (x-1)^{9/2}}{108
   \sqrt{3}}+O\left((x-1)^{11/2}\right)$$
Make $x=2$ and obtain
$$\frac {1207 \sqrt 3 } { 2268}\approx 0.9218$$ More terms you will add to the expansion and closer and closer you will approach to the true value which is $\approx 0.8969$.
