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To evaluate:
$I = \int \frac{dx}{\sqrt{x} + \sqrt{1-x}}$

Multiply the numerator and denominator by $(\sqrt{x} - \sqrt{1-x})$:
$I = \int \frac{\sqrt{x} - \sqrt{1-x}}{(\sqrt{x} + \sqrt{1-x})(\sqrt{x} - \sqrt{1-x})}\,dx = \int \frac{\sqrt{x} - \sqrt{1-x}}{x - (1-x)}\,dx = \int \frac{\sqrt{x} - \sqrt{1-x}}{2x - 1}\,dx$

Let $x = \sin^2 \theta$, so $dx = 2\sin\theta\cos\theta\,d\theta$:
Then $\sqrt{x} = \sin\theta$, $\sqrt{1-x} = \cos\theta$, and $2x - 1 = 2\sin^2\theta - 1 = -\cos 2\theta$.

Substituting these into the integral:
$I = \int \frac{\sin\theta - \cos\theta}{-\cos 2\theta} \cdot 2\sin\theta\cos\theta\,d\theta$
$I = \int \frac{\cos\theta - \sin\theta}{\cos^2\theta - \sin^2\theta} \cdot \sin 2\theta\,d\theta$
$I = \int \frac{\cos\theta - \sin\theta}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)} \cdot \sin 2\theta\,d\theta$
$I = \int \frac{\sin 2\theta}{\cos\theta + \sin\theta}\,d\theta$

Notice that $(\sin\theta + \cos\theta)^2 = 1 + \sin 2\theta \implies \sin 2\theta = (\sin\theta + \cos\theta)^2 - 1$.
Therefore,
$I = \int \left( (\sin\theta + \cos\theta) - \frac{1}{\sin\theta + \cos\theta} \right) d\theta$

Alternatively, let $u = \sin\theta - \cos\theta$, then $du = (\cos\theta + \sin\theta)d\theta$ and $u^2 = 1 - \sin 2\theta \implies \sin 2\theta = 1 - u^2$:
$I = \int (\sin\theta + \cos\theta)\,d\theta - \int \frac{1}{\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)}\,d\theta$
$I = \sin\theta - \cos\theta - \frac{1}{\sqrt{2}}\ln\left|\sec\left(\theta - \frac{\pi}{4}\right) + \tan\left(\theta - \frac{\pi}{4}\right)\right| + C$

More simply, in terms of $x$:
Using the substitution $2x - 1 = \sin t$, $2dx = \cos t\,dt$:
$I = \sqrt{x(1-x)} + \frac{1}{2}\arcsin(2x-1) + C$