1. **State the problem:** We need to find the integral with respect to $x$ of the function $$\frac{1}{\pi} \arccos \left(\frac{r-x}{r}\right).$$ 2. **Recall the formula and rules:** The integral of $\arccos(u)$ with respect to $u$ is $$\int \arccos(u) \, du = u \arccos(u) - \sqrt{1-u^2} + C.$$ We will use substitution to handle the inner function $\frac{r-x}{r}$. 3. **Substitution:** Let $$u = \frac{r-x}{r} = 1 - \frac{x}{r}.$$ Then, $$du = -\frac{1}{r} dx \implies dx = -r \, du.$$ 4. **Rewrite the integral:** $$\int \frac{1}{\pi} \arccos\left(\frac{r-x}{r}\right) dx = \frac{1}{\pi} \int \arccos(u) (-r) du = -\frac{r}{\pi} \int \arccos(u) du.$$ 5. **Integrate using the formula:** $$-\frac{r}{\pi} \left(u \arccos(u) - \sqrt{1-u^2}\right) + C = -\frac{r}{\pi} \left(\left(1 - \frac{x}{r}\right) \arccos\left(1 - \frac{x}{r}\right) - \sqrt{1 - \left(1 - \frac{x}{r}\right)^2}\right) + C.$$ 6. **Simplify the square root term:** $$1 - \left(1 - \frac{x}{r}\right)^2 = 1 - \left(1 - \frac{2x}{r} + \frac{x^2}{r^2}\right) = \frac{2x}{r} - \frac{x^2}{r^2} = \frac{x}{r} \left(2 - \frac{x}{r}\right).$$ 7. **Final answer:** $$\boxed{\int \frac{1}{\pi} \arccos \left(\frac{r-x}{r}\right) dx = -\frac{r}{\pi} \left(\left(1 - \frac{x}{r}\right) \arccos \left(1 - \frac{x}{r}\right) - \sqrt{\frac{x}{r} \left(2 - \frac{x}{r}\right)}\right) + C}.$$