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So I think I know how to solve the question I ve been asked however I would like to clarify some doubts I have. So I have to check if $f(x)=\frac{1}{x(1-x)}$ is integrable on $(0,1)$. One could say that $f(x)\geq\frac{1}{x}$ which is clearly divergent since $\log(x)$ escapes to minus infinity near zero. However, I am not sure If I can use the primitive of $\frac{1}{x}$ at all because as I just said $\frac{1}{x}$ is not integrable on (0,1) so maybe the primitive is not defined somewhere? The second approach I had was to substitute $\frac{1}{t}=x$ so potential integral would be $\int_0^1 \frac{1}{x(1-x)}dx = -\int_1^{\infty} \frac{x^2}{x-1}dx$ and if this is integrable then $\sum_{n=1}^\infty\frac{n^2}{n-1}$ is finite but obviously it cannot converge since $\frac{n^2}{n-1}\nrightarrow 0 \ \text{as} \ n\to \infty$.

I just want someone to clarify what is "ok" to do (and why), I got a bit confused.

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  • $\begingroup$ (1) What do you mean "maybe primitive is not defined somewhere?" $\frac1x$ is defined for all nonzero $x$. $\tag*{}$ (2) Check your $x=\frac1t$ substitution; it should be $-\int_1^\infty \frac{1}{\frac1t(1-\frac1t)}d\frac1t$, and it seems you forgot the $d\frac1t(=-\frac1{t^2}\,dt)$ part. $\endgroup$ Jun 29 '18 at 18:17
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$\int_0^1 \frac{1}{x(1-x)}dx$

$ = \int_0^1(\frac{1}{x}+\frac{1}{1-x})dx $

$=\int_0^{1}\frac{1}{x}dx+\int_{0}^1\frac{1}{1-x}dx$, which tends to $+\infty$. Therefore, given integral is divergent.

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