Question:

Step 1: Differentiate the given functional equation with respect to $x$. We are given that $xf(x) = x + \int_{1}^{x} f(t)\,dt$. Differentiating both sides with respect to $x$: $$\frac{d}{dx}[xf(x)] = \frac{d}{dx}\left[x + \int_{1}^{x} f(t)\,dt\right]$$ Using the product rule on the left side and the fundamental theorem of calculus on the right side: $$f(x) + xf'(x) = 1 + f(x)$$ Step 2: Simplify to find $f'(x)$. Subtracting $f(x)$ from both sides: $$xf'(x) = 1$$ Therefore: $$f'(x) = \frac{1}{x}$$ Step 3: Integrate to find the general form of $f(x)$. Integrating both sides with respect to $x$: $$f(x) = \ln x + C$$ where $C$ is a constant of integration. Step 4: Use the initial condition to find the constant $C$. Substitute $x = 1$ into the original equation: $$1 \cdot f(1) = 1 + \int_{1}^{1} f(t)\,dt = 1 + 0 = 1$$ So $f(1) = 1$. Now substitute $x = 1$ into $f(x) = \ln x + C$: $$f(1) = \ln 1 + C = 0 + C = 1$$ Therefore, $C = 1$. Step 5: Write the explicit form of $f(x)$. $$f(x) = \ln x + 1$$ Step 6: Evaluate $f(e^k)$ for the sum. Substitute $x = e^k$ into $f(x)$: $$f(e^k) = \ln(e^k) + 1 = k + 1$$ Step 7: Calculate the sum $\sum_{k=1}^{10} f(e^k)$. $$\sum_{k=1}^{10} f(e^k) = \sum_{k=1}^{10} (k+1)$$ Separating the sum: $$\sum_{k=1}^{10} (k+1) = \sum_{k=1}^{10} k + \sum_{k=1}^{10} 1$$ Using the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$: $$\sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55$$ Therefore: $$\sum_{k=1}^{10} f(e^k) = 55 + 10 = 65$$ The answer is **Option 3: 65**.