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.
Note that for $1>\varepsilon>0$ we have
$$\begin{align} \int_{1+\varepsilon}^2\frac{1}{\sqrt{x^4-x}}\,dx&=\int_{1+\varepsilon}^2\frac{1}{\sqrt{x(x^2+x+1)(x-1)}}\,dx\\\\ &\le \frac1{\sqrt {3}}\int_{1+\varepsilon}^2\frac1{\sqrt{x-1}}\,dx\\\\ &=\frac2{\sqrt{3}}\left(1-\sqrt{\varepsilon}\right) \end{align}$$
$$\int_1^2\frac1{\sqrt{x^4-x}}<\int_1^2\frac1{\sqrt{x^4-x^2}}=\frac\pi3$$
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 $\approx 0.8969$.
