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

$\frac{1}{2}$

The correct answer is Option (3) → $\frac{1}{2}$

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

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

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

eq. (1) + eq. (2)

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

$2I=\int\limits^{2}_{1}1dx⇒2I=[2-1]$

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