Question

${\int }_1^2\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}dx$

Answer 1

${\int }_1^2\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}dx$

$\Rightarrow {\int }_1^2\frac{(\sqrt{x}+\sqrt{3-x})}{(\sqrt{x}+\sqrt{3-x})}dx={\int }_1^{2}1.dx$

$=x{|}_1^2=2-1$

$=1$