Value of $\int_2^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{5-x}}dx$ is

Given:

$\int_2^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{5-x}}dx$

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The King Property of definite integrals states:

$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$

Here $a = 2$ and $b = 3$, so $a + b = 5$.

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Let $I = \int_2^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{5-x}}dx$ ... (1)

Replacing $x$ with $(5-x)$:

$I = \int_2^3 \frac{\sqrt{5-x}}{\sqrt{5-x} + \sqrt{5-(5-x)}}dx$

$I = \int_2^3 \frac{\sqrt{5-x}}{\sqrt{5-x} + \sqrt{x}}dx$ ... (2)

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Adding equations (1) and (2):

$I + I = \int_2^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{5-x}}dx + \int_2^3 \frac{\sqrt{5-x}}{\sqrt{x} + \sqrt{5-x}}dx$

$2I = \int_2^3 \frac{\sqrt{x} + \sqrt{5-x}}{\sqrt{x} + \sqrt{5-x}}dx$

$2I = \int_2^3 1 \, dx$

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$2I = [x]_2^3$

$2I = 3 - 2$

$2I = 1$

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

Therefore, the value of the integral is $\frac{1}{2}$.