The value of $\int\limits^{2}_{1}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx= $ is :

$\frac{1}{2}$

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

$I=\int\limits^{2}_{1}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$   ...(1)

$I=\int\limits^{2}_{1}\frac{\sqrt{1+2-x}}{\sqrt{3-(1+2-x)}+\sqrt{1+2-x}}dx$

$I=\int\limits^{2}_{1}\frac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}dx$   ...(2)

eq. (1) + eq. (2)

$2I=\int\limits^{2}_{1}1dx=1$

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