1. **Problem statement:** Given the function $f(x) = x^2 \ln(3 + 2x)$, we want to find the coefficients $c_n$ of the power series representation of its integral $\int f(x) \, dx = \sum_{n=0}^\infty c_n x^n$, the radius of convergence $R$, the interval of convergence endpoints $a$ and $b$, and determine convergence at these endpoints. 2. **Step 1: Express $\ln(3 + 2x)$ as a power series.** Recall the Taylor series for $\ln(1 + t)$ around $t=0$ is: $$\ln(1+t) = \sum_{k=1}^\infty (-1)^{k+1} \frac{t^k}{k}$$ Set $t = \frac{2x}{3}$ so that: $$\ln(3 + 2x) = \ln\left(3\left(1 + \frac{2x}{3}\right)\right) = \ln 3 + \ln\left(1 + \frac{2x}{3}\right) = \ln 3 + \sum_{k=1}^\infty (-1)^{k+1} \frac{(2x/3)^k}{k}$$ 3. **Step 2: Multiply by $x^2$ to get $f(x)$ series:** $$f(x) = x^2 \ln(3 + 2x) = x^2 \ln 3 + x^2 \sum_{k=1}^\infty (-1)^{k+1} \frac{(2/3)^k}{k} x^k = x^2 \ln 3 + \sum_{k=1}^\infty (-1)^{k+1} \frac{(2/3)^k}{k} x^{k+2}$$ 4. **Step 3: Integrate term-by-term to find $\int f(x) dx$:** Integrate each term: $$\int f(x) dx = \int x^2 \ln 3 \, dx + \int \sum_{k=1}^\infty (-1)^{k+1} \frac{(2/3)^k}{k} x^{k+2} \, dx$$ The first integral: $$\int x^2 \ln 3 \, dx = \ln 3 \int x^2 \, dx = \ln 3 \cdot \frac{x^3}{3} + C$$ The second integral: $$\sum_{k=1}^\infty (-1)^{k+1} \frac{(2/3)^k}{k} \int x^{k+2} \, dx = \sum_{k=1}^\infty (-1)^{k+1} \frac{(2/3)^k}{k} \cdot \frac{x^{k+3}}{k+3} + C$$ 5. **Step 4: Write the power series for $\int f(x) dx$ as $\sum_{n=0}^\infty c_n x^n$.** Note the powers of $x$ in the integral series start at $x^3$ from the first term $\ln 3 \frac{x^3}{3}$ and from the sum at $x^{k+3}$ with $k \geq 1$. So: - For $n=0,1,2$, $c_n = 0$ because no terms of degree less than 3. - For $n=3$, the coefficient comes from $\ln 3 \frac{1}{3}$ (from the first integral term) plus the $k=0$ term in the sum which does not exist, so only the first term contributes. - For $n \geq 4$, $c_n$ comes from the sum with $k = n-3$: $$c_n = (-1)^{(n-3)+1} \frac{(2/3)^{n-3}}{(n-3)(n)}$$ 6. **Step 5: Write explicit formulas for $c_1, c_2, c_3, c_n$:** - $c_1 = 0$ - $c_2 = 0$ - $c_3 = \frac{\ln 3}{3}$ - For $n \geq 4$: $$c_n = (-1)^{n-2} \frac{(2/3)^{n-3}}{(n-3) n}$$ 7. **Step 6: Find the radius of convergence $R$.** The radius of convergence of the original series for $\ln(1+t)$ is $|t| < 1$. Since $t = \frac{2x}{3}$, the radius in $x$ is: $$\left| \frac{2x}{3} \right| < 1 \implies |x| < \frac{3}{2}$$ Integration does not change the radius of convergence, so: $$R = \frac{3}{2}$$ 8. **Step 7: Interval of convergence endpoints $a$ and $b$.** $$a = -\frac{3}{2}, \quad b = \frac{3}{2}$$ 9. **Step 8: Check convergence at endpoints.** At $x = a = -\frac{3}{2}$, substitute into $t = \frac{2x}{3} = -1$. The original series for $\ln(1+t)$ converges conditionally at $t=-1$ (alternating harmonic series). Multiplying by $x^2$ and integrating termwise preserves convergence. Hence, the series converges at $x=a$. At $x = b = \frac{3}{2}$, $t=1$. The series for $\ln(1+t)$ diverges at $t=1$ (harmonic series without alternating sign). Therefore, the series diverges at $x=b$. **Final answers:** $c_1 = 0$ $c_2 = 0$ $c_3 = \frac{\ln 3}{3}$ For $n \geq 4$: $$c_n = (-1)^{n-2} \frac{(2/3)^{n-3}}{(n-3) n}$$ Radius of convergence: $$R = \frac{3}{2}$$ Interval of convergence: $$a = -\frac{3}{2}, \quad b = \frac{3}{2}$$ Convergence at endpoints: - At $x=a$: 1 (converges) - At $x=b$: 0 (diverges)