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Let $I=\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$                ...(1)

It is known that, $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$                   ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$

$\Rightarrow 2 I=\int_{0}^{a} 1 d x$

$\Rightarrow 2 I=[x]_{0}^{a}$

$\Rightarrow 2 I=a$

$\Rightarrow I=\frac{a}{2}$