$\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$ is equal to

Let $I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$ ... (1)

Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$, replace $x$ with $(a-x)$:

$I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{a-(a-x)}} dx$

Since $a-(a-x) = x$:

$I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx$ ... (2)

Adding equations (1) and (2):

$I + I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx + \int_0^a \frac{\sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} dx$

$2I = \int_0^a \left[\frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}}\right] dx$

$2I = \int_0^a 1 \, dx$

$2I = [x]_0^a$

$2I = a - 0$

$2I = a$

$I = \dfrac{a}{2}$

Therefore, $\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx = \dfrac{a}{2}$