Integrating $\int_0^3 \frac{\sqrt x}{\sqrt x + \sqrt{3 - x}} \,dx$ This integral is unique for me and I was quite wondering how an integral below would result into "$3/2$". 
Show that 
$$
\int_0^3 \frac{\sqrt x}{\sqrt x + \sqrt{3 - x}} \ dx =\frac{3}{2}
$$

Substitute with $y=3-x$ gives
$$I=-\int_3^0 \frac{\sqrt{3-y}}{\sqrt y +\sqrt{3-y}} dy=\int_0^3  \frac{\sqrt{3-y}}{\sqrt y +\sqrt{3-y}} dy$$
So
$$2I=3$$
