**Toolbox:**

- $\int \limits_0^a f(x) dx=\int \limits_0^a f(a-x) dx$

$\int\limits_{0}^{3}\large\frac{\sqrt{x}dx}{\sqrt{x}+\sqrt{3-x}}$

Step 1:

$\int\limits_{0}^{3}\large\frac{\sqrt{x}dx}{\sqrt{x}+\sqrt{3-x}}$$=I$

$\qquad=\int\limits_{0}^{3}\large\frac{\sqrt{3-x}dx}{\sqrt{3-x}+\sqrt{3-(3-x)}}$

$\qquad=\int\limits_{0}^{3}\large\frac{\sqrt{3-x}dx}{\sqrt{3-x}+\sqrt{x}}$

Step 2:

$\therefore 2I=\int \limits _0^3 \bigg[\large\frac{\sqrt x}{\sqrt x +\sqrt 3 -x}+\frac{\sqrt {3-x}}{\sqrt {3-x}+\sqrt x}\bigg]$$dx$

$\qquad=\int \limits_0^3 dx=x \bigg]_0^3=3$

$=>I=\large\frac{3}{2}$