$\int\limits_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$ is equal to

$\frac{a}{2}$

The correct answer is Option (2) → $\frac{a}{2}$

Let $I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx$

Use $\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $f(x)=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}$

$f(a-x)=\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}$

$f(x)+f(a-x)=1$

$\Rightarrow\ 2I=\int_{0}^{a}1\,dx=a$

Therefore,$ I=\frac{a}{2}$

Final Answer: $\frac{a}{2}$