$\int\limits_1^2\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$ is equal to

$\frac{1}{2}$

The correct answer is Option (2) → $\frac{1}{2}$

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

Use property $\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx$ Here $a=1,\ b=2 \Rightarrow a+b=3$

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

Add both:

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

∴ $I=\frac{1}{2}$

Final Answer: $\frac{1}{2}$