Find the value of the integral !

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Find the value of the integral !

Mathematics
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\[\int\limits_{1}^{2}\sqrt{\frac{ 2-x }{ x-1 }}\]
Studied the convergence and found that it converges
now I must find to what it converges

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\[u=\sqrt{x-1} \\ u^2=x-1 \\ 2u du=dx \\ \int\limits_0^1 \frac{\sqrt{2-(u^2+1)}}{u} 2 u du \\ 2 \int\limits_0^1 \sqrt{1-u^2} du\] try this integral
you actually can find that using just a geometrical approach it is just a quarter of a circle with radius 1 multiplied by 2
i've run a simple numerical on this and it tells me that the answer is 1 but that you need to be very careful about the start of the interval [which is why i ran the numerical in first place]. this makes me wonder if this is really to be approached as \( \lim_{a \rightarrow 1+} \ \int_{a}^{2} \sqrt{\frac{2-x}{x-1}}\). no idea if this is helpful, but hope it does help.
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