Show an improper integral $\int_0^1 \frac{dx}{\sqrt{x^3-x}}$ converges I would like to show that the following improper integral converges, but it's been a while since I've done this sort of calculus and I'm drawing a blank:
$$
\int_0^1 \frac{dx}{\sqrt{x^3-x}}
$$
My first thought was to try and simplify the expression under the radical. I tried to factor out $\sqrt{x}$ and use partial fractions, but that didn't really lead anywhere. A similar thing happened when I tried to factor out $x$. Next I thought about ways I could say that the integral converged without actually evaluating it, but again, I didn't really get anywhere with any of the standard integration techniques. Any thoughts?
 A: When $x \in (0, 1/\sqrt{2})$, we have $x^3 - x/2 < 0$. It follows that
$$
\frac{1}{\sqrt{x-x^3}} < \frac{1}{\sqrt{x - x/2}} = \frac{\sqrt{2}}{\sqrt{x}}
$$
on the same interval. Since $\int_0^1 dx/\sqrt{x} $ converges, your integral also converges on $(0, 1/\sqrt{2})$.
For the other side of the improper integral, use the following similar comparison:
$$
\frac{1}{\sqrt{x-x^3}} < \frac{\sqrt{2}}{\sqrt{1-x}}
$$
on $x \in \left(\frac{-1+\sqrt{5}}{2}, 1\right)$.
A: Let $I = \displaystyle \int_0^1 \dfrac{dx}{\sqrt{x-x^3}}$.
$x = t^2 \implies I = \displaystyle \int_0^1 \dfrac{dx}{\sqrt{x-x^3}} = \displaystyle \int_0^1 \dfrac{2t dt}{\sqrt{t^2-t^6}}= 2 \int_0^1 \dfrac{dt}{\sqrt{1-t^4}}$
$$\dfrac1{\sqrt{1-t^4}} = \dfrac1{\sqrt{1-t} \sqrt{1+t} \sqrt{1+t^2}} \leq \dfrac1{\sqrt{1-t}}$$
for all $t \in [0,1]$.
Hence, $$I \leq 2 \int_0^1 \dfrac{dt}{\sqrt{1-t}} = - \left. 2 \dfrac{\left(1-t\right)^{1/2}}{1/2} \right\vert_0^1 = 4$$
A: $$
\int_0^1\frac{dx}{\sqrt{x^3-x}} = \int_0^1 \frac{dx}{i\sqrt{x-x^3}} = \frac{1}{i}\int_0^1 \frac{dx}{\sqrt{x-x^3}}
$$
Since $x-x^3\geq0\;\forall\;x\in[0,1]$, the integral clearly converges.
