Question: This integral is unique for me and I was quite wondering how an integral below would result into "$3/2$". 

Show that 
$$
\int\_0^3 \frac{\sqrt x}{\sqrt x + \sqrt{3 - x}} \ dx =\frac{3}{2}
$$

Answer: 
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

$$2I=3$$
Personally, I'd multiply by $\sqrt{x}-\sqrt{3-x}$ and integrate by parts. And I'd be careful about the domain (separate it in two parts, one $x<\frac{3}{2}$ and the other $x>\frac{3}{2}$)

But I am sure there are plenty of other tricks applicable (variable change for instance)