$\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx$ is equal to

The integral to evaluate is:

$\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx$

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For definite integrals, the following property holds:

$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$

Here $a = 1$ and $b = 2$, so $a + b = 3$.

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Let $I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx$

Replacing $x$ with $3-x$:

$I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3 - (3-x)} + \sqrt{3-x}} dx$

$I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} dx$

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Adding both expressions for $I$:

$I + I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx + \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3-x} + \sqrt{x}} dx$

$2I = \int_{1}^{2} \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{3 - x} + \sqrt{x}} dx$

$2I = \int_{1}^{2} 1 \, dx$

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$2I = [x]_{1}^{2}$

$2I = 2 - 1$

$2I = 1$

$I = \frac{1}{2}$

Therefore, the integral equals $\frac{1}{2}$.