Be careful with that domain!

For integral points, $\left \{x+1\right \} =0$

So we must have, $2x>x^2\implies x(2-x)>0$ which is satisfied by $\boxed{x=1}$

Here we go,

$\large f(x) = \dfrac{1}{\sqrt{ \{x+1\} - x^2 + 2x }}$

So we know $\{x+1\} - x^2 + 2x > 0$

$\implies$ $\{x+1\} > x^2 - 2x$

Again we know that,

$\implies$ $0 \leq \{x+1\} < 1$

Hence it is clear that

$\implies$ $x^2-2x<1$

$\implies x^2-2x + 1 <2$

$\implies (x-1)^2<2$

$\implies (x-1)^2-\sqrt{2}^2<0$

$\implies (x-1-\sqrt{2})(x-1+\sqrt{2}<0$

So $x \in (1-\sqrt{2} , 1+\sqrt{2})$

Now see that we have $x= 0 , 1, 2$ as integers

But by observation $x=0$ and $x=2$ makes $f(x)$ undefined.

$\implies$ Only integral $x$ which satisfies is $x=1$

So $\boxed{1}$ integer points the domain of the given function contains.