Let's analyze the integral: \[ \int_{0}^{\infty} \frac{\ln(x+1)}{3-x} \, dx \] We are asked to **give every reason** why this is an improper integral. --- ### **Step-by-Step Explanation** #### 1. **Infinite Interval** The upper limit of integration is \(\infty\). This makes the integral improper because the interval of integration is unbounded. - **Reason 1:** The upper limit is infinite (\(\infty\)), so the integral is improper at the upper limit. #### 2. **Singularity in the Integrand** Look at the denominator \(3-x\): - When \(x = 3\), the denominator becomes zero, so the integrand blows up (goes to infinity). Since \(x = 3\) is within the interval of integration (\(0 \leq x < \infty\)), the integrand is undefined at \(x = 3\). - **Reason 2:** The integrand has a vertical asymptote (division by zero) at \(x = 3\), which is within the interval of integration. #### 3. **Check Lower Limit** At \(x = 0\), \(\ln(x+1) = \ln(1) = 0\), so the integrand is 0. No problem here. #### 4. **Check for any other singularities** There are no other singularities between 0 and \(\infty\) except at \(x = 3\). --- ### **Final Answer** **The integral \(\int_{0}^{\infty} \frac{\ln(x+1)}{3-x} dx\) is improper for the following reasons:** 1. **The upper limit of integration is infinite (\(\infty\)).** 2. **The integrand has a vertical asymptote at \(x = 3\), where the denominator becomes zero, and this point lies within the interval of integration \([0, \infty)\).** --- **Summary Table:** | Reason | Where? | Why? | |----------------------------------|------------|---------------------------------------------------| | Infinite limit | \(x \to \infty\) | The interval is unbounded | | Division by zero (asymptote) | \(x = 3\) | Denominator \(3-x = 0\) | --- **Thus, the given integral is improper for both of the above reasons.**

Let's analyze the integral carefully and understand why it is classified as an improper integral. --- ### **Given Integral:** \[ I = \int_{0}^{\infty} \frac{\ln(x+1)}{3 - x} \, dx \] --- ### **Step 1: Recognize the features of the integral that make it improper.** An integral is called *improper* if: - It has an **infinite limit of integration** (either upper or lower limit is \(\infty\) or \(-\infty\)) - The integrand has a **discontinuity or singularity** within the interval of integration (where the integrand becomes unbounded or undefined) --- ### **Step 2: Check the limits of integration** - **Lower limit:** \(x=0\) - **Upper limit:** \(x=\infty\) **Observation:** The upper limit is \(\infty\), which makes the integral *improper at the upper limit.* **Reason 1:** Since the upper limit of the integral is infinity, the integral extends over an unbounded interval. This is a classic case of an improper integral because we need to evaluate the limit: \[ \lim_{t \to \infty} \int_0^t \frac{\ln(x+1)}{3 - x} \, dx \] --- ### **Step 3: Check the integrand for singularities within the interval** The integrand is: \[ f(x) = \frac{\ln(x+1)}{3 - x} \] - **Potential issue 1:** The denominator \(3 - x\) becomes zero at \(x = 3\). **Analysis:** - At \(x=3\), the denominator is zero. - The numerator \(\ln(3+1) = \ln 4\), which is finite. - Therefore, the integrand approaches infinity at \(x=3\). **Observation:** Since \(x=3\) lies in the interval \([0, \infty)\), the integrand has a **vertical asymptote** at \(x=3\). **Reason 2:** The integrand has a **discontinuity within the interval** at \(x=3\), where it tends to infinity. --- ### **Step 4: Is the integral improper at \(x=3\)?** Yes. Because: - The integrand becomes unbounded at \(x=3\). - To properly evaluate the integral, it must be split at \(x=3\), and limits must be taken approaching 3 from the left and right. --- ### **Step 5: Summary of reasons why the integral is improper** | **Reason** | **Explanation** | |--------------|----------------| | **Infinite upper limit** | The upper limit \( \infty \) makes the integral improper because we need to evaluate a limit as \( t \to \infty \) of a definite integral over \([0, t]\). | | **Singularity at \(x=3\)** | The integrand's denominator zeroes out at \(x=3\), causing a vertical asymptote within the interval \([0, \infty)\). This makes the integral improper because the integrand is unbounded at that point. | --- ### **Final answer:** **The integral \(\int_{0}^{\infty} \frac{\ln(x+1)}{3 - x} \, dx\) is improper because:** 1. **The upper limit of integration is infinite (\(\infty\)), which makes the integral improper at the upper limit.** 2. **The integrand has a discontinuity (singularity) at \(x=3\), where the denominator \(3 - x\) becomes zero, causing the integrand to tend to infinity within the interval of integration.** --- **In conclusion**, both reasons contribute to the integral being classified as improper, and proper evaluation would involve splitting the integral at \(x=3\) and taking appropriate limits at both the infinity and the singular point.