1. **Problem Statement:** We are given a profit function for a production division: $$P(x) = \frac{500 \ln(x + 1)}{(x + 1)^2}$$ where $x$ is in hundreds of units and $P(x)$ is in thousands of dollars. We need to find the average profit over the interval from $x=0$ to $x=10$ hundred units using integral calculus. 2. **Formula for Average Value of a Function:** The average value $\bar{P}$ of a continuous function $P(x)$ over the interval $[a,b]$ is given by: $$\bar{P} = \frac{1}{b - a} \int_a^b P(x) \, dx$$ Here, $a=0$ and $b=10$. 3. **Set up the integral:** $$\bar{P} = \frac{1}{10 - 0} \int_0^{10} \frac{500 \ln(x + 1)}{(x + 1)^2} \, dx = \frac{1}{10} \int_0^{10} \frac{500 \ln(x + 1)}{(x + 1)^2} \, dx$$ 4. **Simplify the constant factor:** $$\bar{P} = 50 \int_0^{10} \frac{\ln(x + 1)}{(x + 1)^2} \, dx$$ 5. **Substitution:** Let $t = x + 1$, so when $x=0$, $t=1$ and when $x=10$, $t=11$. Also, $dx = dt$. The integral becomes: $$\int_1^{11} \frac{\ln t}{t^2} \, dt$$ 6. **Integration by parts:** Let: - $u = \ln t \implies du = \frac{1}{t} dt$ - $dv = t^{-2} dt \implies v = -t^{-1}$ Then: $$\int \frac{\ln t}{t^2} dt = uv - \int v \, du = -\frac{\ln t}{t} - \int -\frac{1}{t} \cdot \frac{1}{t} dt = -\frac{\ln t}{t} + \int \frac{1}{t^2} dt$$ 7. **Evaluate the remaining integral:** $$\int \frac{1}{t^2} dt = \int t^{-2} dt = -t^{-1} + C = -\frac{1}{t} + C$$ 8. **Combine results:** $$\int \frac{\ln t}{t^2} dt = -\frac{\ln t}{t} - \frac{1}{t} + C = -\frac{\ln t + 1}{t} + C$$ 9. **Evaluate definite integral from 1 to 11:** $$\int_1^{11} \frac{\ln t}{t^2} dt = \left[-\frac{\ln t + 1}{t}\right]_1^{11} = \left(-\frac{\ln 11 + 1}{11}\right) - \left(-\frac{\ln 1 + 1}{1}\right)$$ Since $\ln 1 = 0$: $$= -\frac{\ln 11 + 1}{11} + 1 = 1 - \frac{\ln 11 + 1}{11}$$ 10. **Calculate the average profit:** $$\bar{P} = 50 \times \left(1 - \frac{\ln 11 + 1}{11}\right) = 50 \left(1 - \frac{\ln 11 + 1}{11}\right)$$ This is the average profit in thousands of dollars over the interval from 0 to 10 hundred units. **Final answer:** $$\boxed{\bar{P} = 50 \left(1 - \frac{\ln 11 + 1}{11}\right) \text{ thousand dollars}}$$ This completes the solution.