Determine the primitives Determine the primitives of $ f:(0,\infty )\rightarrow \mathbb{R},f(x)=\frac{\sqrt{[x]}+\sqrt{\left \{ x \right \}}}{\sqrt{x}} $ , where $ \left \{ x \right \} $ represents the fractional part and $ [x] $, the floor of $ x $.
I didn't find a suitable method to integrate the function.
 A: Hint 1: When dealing with the floor function, it is often useful to check how it behaves on intervals where the floor function is constant. Also notice that $\{x\} = x-[x]$. So for $x\in [n,n+1)$, $n \in \mathbb{N}$ we have $[x] = n$ and the function simplifies to 
$$
f(x) = \frac{\sqrt{n}+\sqrt{x-n}}{\sqrt{x}}.
$$
Then you want to integrate 
$$\int f(x)\,dx=\int \frac{\sqrt{n}+\sqrt{x-n}}{\sqrt{x}}\,dx = \sqrt{n}\int\frac{1}{\sqrt{x}}\,dx+\int\sqrt{1-\frac{n}{x}}\,dx$$
and use the fact that $(0,\infty)$ is union of all such intervals $[n,n+1)$ (excluding $0$).
Hint 2:
The second integral in the expression above is a bit tricky. Consider case $n=1$ and substitution $t^2=1-\frac{1}{x}$. Then you have $x \in [1,2)$ and $t\in \left[0,\frac{1}{\sqrt{2}}\right)$. You can write 
\begin{align}
x&=\frac{1}{1-t^2}\\
dx&=\frac{2t}{(1-t^2)^2}dt
\end{align}
Finally plugging this into the integral
$$
\int\sqrt{1-\frac{1}{x}}\,dx = \int \frac{2t^2}{(1-t^2)^2}\,dt
$$ 
and you should be able to solve this using parcial fractions (be careful with the signs, especially because $t-1< 0$). You can generalize this to a generic $n$, it will just get messy.
