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I solved this question and was sure about my answer that it converges but the answer was it diverges, with no explanation,
I still don't get what's my mistake, here's what I did:

I wanted to use the limit comparison test, $\displaystyle \frac {1}{x^{0.5}(1-x)^2}=\frac {1}{x^{0.5}(1-2x+x^2)}=\frac {1}{x^{0.5}-2x^{1.5}+x^{2.5}}$, at $x\to 0$ , this function behaves like $\displaystyle \frac {1}{x^{0.5}}$.
$\displaystyle \lim_{x\to0} \frac {x^{0.5}}{x^{0.5}-2x^{1.5}+x^{2.5}}$ then I divided by $x^{0.5}$ and got $\displaystyle \lim_{x\to0} \frac {1}{1-2x^{}+x^2}=1$.

And the integral $\displaystyle \int_0^1 \frac{dx}{x^{0.5}}$ converges, then my integral should converge too.

Now I'm not sure where is my mistake, I would appreciate any help.

Thanks in advance.

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You are right about $x=0$. It's what happens at $x=1$ that makes your integral diverge.

To be more precise: your integral converges if and only if both integrals$$\int_0^{1/2}\frac1{\sqrt x(1-x)^2}\,\mathrm dx\quad\text{and}\quad\int_{1/2}^1\frac1{\sqrt x(1-x)^2}\,\mathrm dx$$converge. Your computations show indeed that the first one converges. But the second one diverges.

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  • $\begingroup$ ohh thanks, I didn't notice that, appreciate it! $\endgroup$ – Pwaol Jan 31 at 19:00
  • $\begingroup$ If I take $\frac {1}{(1-x)^2}$ to compare the second one, then substitute $t=1-x$ in the integral, and get $\int^1_{1/2} \frac {1}{t^2}$, that would be a validate proof that the second integral diverges? $\endgroup$ – Pwaol Jan 31 at 19:12
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    $\begingroup$ @Pwaol Yes, that is correct. $\endgroup$ – José Carlos Santos Jan 31 at 19:16

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