I want to evaluate the following integral:
$$ \int_{1/2}^1 \frac{1}{x \sqrt{(1-x)x}} dx$$
I would compute this as
$$ \frac{2(x-1)}{\sqrt{(1-x)x}}\biggr\rvert_{1/2}^1 $$
which is not defined at $1$.
However, Mathematica evaluates the definite integrals as $2$.
I can see how one would obtain approximately 2 by evaluating it close to 1, but is it correct?
In other words: Is $2$ the correct answer, or is the integral really not defined?
$$\frac{2(x-1)}{\sqrt{(1-x)x}} = -\frac{2(\sqrt{1-x})^2}{\sqrt{(1-x)x}} = -2\sqrt{\frac{1-x}{x}}$$
which should be easy to evaluate at $x=1$.
This is an improper integral and should indeed be interpreted as $$ \lim_{t\to1}\int_{1/2}^t\frac{1}{x\sqrt{x(1-x)}}\,dx $$ You have correctly computed the antiderivative, so you get $$ \lim_{t\to 1}-\frac{2(1-x)}{\sqrt{x(1-x)}}\,\bigg|_{1/2}^t= \lim_{t\to 1}\left(-\frac{2(1-t)}{\sqrt{t(1-t)}}+\frac{1}{\sqrt{1/4}}\right) $$ However, $$ -\frac{2(1-t)}{\sqrt{t(1-t)}}=-\frac{2\sqrt{1-t}}{\sqrt{t}} $$ so the limit is indeed $2$.
