I wanted to find the following antiderivative $$\int \frac{dx}{\sqrt{x}\sqrt{1-x}}.$$ Obviously, this does only make sense (when working with real numbers) on the interval $(0,1)$. So essentially we want to find $$\int \frac{dx}{\sqrt{x-x^2}}.$$ In fact, that computation is the same as finding $$\int \frac{dx}{0.25-(x-0.5)^2}.$$ It is easy check that the solution $$\arcsin(2x-1)+c.$$
However, WolframAlpha's answer is a bit different $$\frac{2 \sqrt{x - 1} \sqrt{x} \log(\sqrt{x - 1} + \sqrt{x})}{\sqrt{-(x - 1) x}} + c.$$
However, in the real case (remember $x \in (0,1)$) WolframAlpha's answer does not make sense because we have the square root of a negative number $x-1$.
Does anybody know what is happening there and why WA tries to solve the integral in that way?
The two answers are equivalent, as they differ by a constant. We can show this by showing that the derivative of their difference is zero. As you're already working with WolframAlpha, you might use it to check by simply entering the following:
simplify diff (2sqrt(x-1)sqrt(x)log(sqrt(x-1)+sqrt(x)))/sqrt(x(1-x))-asin(2x-1)
The result is zero. We can also find what the constant is by plugging in a convenient number, like $x=1/2$. We can also do this in WolframAlpha:
(2sqrt(x-1)sqrt(x)log(sqrt(x-1)+sqrt(x)))/sqrt(x(1-x))-asin(2x-1) /. x->1/2
We find that the difference is $-\pi/2$.
On the other hand, while the answers are equivalent, the WolframAlpha answer is more complicated than it might be. But, if you ask WolframAlpha to show its steps, you get exactly the answer that you were looking for: