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I solved this improper integral and I know that the result is correct but I wanted to understand if the theorems I used and the steps I took are actually correct $$ \displaystyle\int\limits_{-1}^1\frac{\sqrt{|x|}}{|x|-1}\,dx $$ I broke the integral into x=0 and since it is an even function I studied the interval [0,1). Subsequently I noticed that the function is negative on the interval and I decided to use the asymptotic comparison criterion. I rewrote the function as $$ \frac{1}{\sqrt{x}(1-\frac{1}{x})}\ $$ and I chose $$ g(x) = 1-\frac{1}{x}$$ I solved lim x-> 1 f(x)/g(x) = 1 and according to the asymptotic comparison criterion g(x) behaves like f(x) in this interval. So I started studying g(x).

I noticed that g(x) is also negative in this interval and I decided to use the asymptotic comparison here too. As the function to compare I chose h(x) = -1/x and since lim x->0 g(x)/h(x) = 1, h(x) behaves like g(x) in this interval.

So since -1/x diverges negatively, g(x) and consequently f(x) also diverge negatively. So I concluded that the integral diverges negatively.

Since I'm not very familiar with it yet, I was wondering if I had followed all the steps correctly or if there was something wrong.

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    $\begingroup$ Seems like the integral does not converge in that domain. Thought the anti derivative of the function in the first integral is $4(\sqrt{x}- \text{arctanh}{\sqrt{x}})$ $\endgroup$ Feb 10 at 12:50

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By Comparison Test

As OP said, the integrand is even and hence $$ \int_{-1}^1 \frac{\sqrt{|x|}}{|x|-1} d x=2 \int_0^1 \frac{\sqrt{x}}{x-1} d x $$ For any $x\in (0,1),$ $$ \frac{\sqrt{x}}{x-1}<\frac{1}{x-1}, $$ and $$ \int_0^1 \frac{1}{x-1} d x=[\ln |x-1|]_0^1=\infty $$

Hence $ \displaystyle \int_{-1}^1 \frac{\sqrt{|x|}}{|x|-1} d x$ is divergent.

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