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: 
Use property $\int\_{a}^{b}f(x)dx$ = $\int\_{a}^{b}f(a+b-x)dx$
**Hint for the standard method:**

The domain of the integrand is $\;0\le x\le 3$. Rewrite it as $\; \dfrac 1{1+\sqrt{\dfrac3x-1}}$ and use the substitution
$$\newcommand\d{\,{\mathrm d\mkern1mu}} u=\sqrt{\dfrac3x-1}\iff x=\frac3{u^2+1},\enspace u\ge 0,\qquad \d x=-\frac{6u}{(u^2+1)^2}\d u$$
So one obtains
$$\int\_0^{+\infty}\frac{6u}{(1+u)(1+u^2)^2}\d u.$$
There remains to use decomposition into partial fractions.