Integral of $\int \frac{1}{\sqrt{x(1-x)}} dx $

First of all, we never have to ask whether we got a correct antiderivative - because we can just take the derivative of our answer and see what we get: $$ \frac d{dx} \big( 2\sin^{-1}(\sqrt x) +C \big) = \frac2{\sqrt{1-(\sqrt x)^2}}\frac d{dx}\sqrt x = \frac2{\sqrt{1-x}} \frac1{2\sqrt x} = \frac1{\sqrt{x(1-x)}}. $$ So you did it right.

There's another way of finding the antiderivative: complete the square inside the square root to see that $$ \frac1{\sqrt{x(1-x)}} = \frac1{\sqrt{1/4 - (x-1/2)^2}} = \frac2{\sqrt{1-(2x-1)^2}}. $$ Therefore, using the substitution $u=2x-1$, \begin{align*} \int \frac1{\sqrt{x(1-x)}} \,dx = \int \frac2{\sqrt{1-(2x-1)^2}} \,dx &= \int \frac1{\sqrt{1-u^2}} \,du \\ &= \sin^{-1} u + C = \sin^{-1} (2x-1) + C. \end{align*} This can also be verified by differentiating.

(Side note: it's possible to check directly that $\sin^{-1}(2x-1)+\pi/2 = 2\sin^{-1}(\sqrt x)$, by taking the cosine of both sides, using $\cos(\theta+\pi/2) = -\sin\theta$ on the left and the double-angle formula $\cos 2\theta = 1-2\sin^2\theta$ on the right. Pretty cool!)