4. Find all real numbers $x$ for which the value of the expression

$$
\sqrt{x^{2}+2 x-3}+\frac{2}{\pi} \cdot \arcsin \frac{x+1}{2}
$$

is an integer.
IV/4. Let's denote $f(x)=\sqrt{x^{2}+2 x-3}+\frac{2}{\pi} \cdot \arcsin \frac{x+1}{2}$. First, let's see for which $x$ the function $f$ is defined. It must hold that $x^{2}+2 x-3 \geq 0$ and $-1 \leq \frac{x+1}{2} \leq 1$. The first condition can be written as $(x+3)(x-1) \geq 0$, from which it follows that $x \leq -3$ or $x \geq 1$. The second condition can be rewritten as $-2 \leq x+1 \leq 2$ or $-3 \leq x \leq 1$. Therefore, the function is defined at $x=-3$ and $x=1$. In the first case, $f(-3)=0+\frac{2}{\pi} \cdot \left(-\frac{\pi}{2}\right)=-1$, and in the second case, $f(1)=0+\frac{2}{\pi} \cdot \frac{\pi}{2}=1$. The only real numbers where the value of the function is an integer are thus $x=-3$ and $x=1$.

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-14.jpg?height=60&width=1631&top_left_y=1695&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-14.jpg?height=52&width=1627&top_left_y=1750&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-14.jpg?height=57&width=1627&top_left_y=1802&top_left_x=223)

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-14.jpg?height=54&width=1627&top_left_y=1852&top_left_x=223)

The function $f$ (or the expression) is defined only for $x=1$ and $x=-3 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

The value of the function $f$ (or the expression) is an integer at $x=1$ and $x=-3 \ldots \ldots$. 1 point each