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 
$$
\int_0^3 \frac{\sqrt x}{\sqrt x + \sqrt{3 - x}} \ dx =\frac{3}{2}
$$

Use property $\int_{a}^{b}f(x)dx$ = $\int_{a}^{b}f(a+b-x)dx$
