Question:

**Step 1: Determine $f'(x)$.** Given the function: $$f(x) = 1 + x + \int_1^x (\ln^2 t + 2 \ln t) \, dt$$ Applying the Fundamental Theorem of Calculus, we differentiate $f(x)$ with respect to $x$: $$f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}\left(\int_1^x (\ln^2 t + 2 \ln t) \, dt\right)$$ $$f'(x) = 0 + 1 + (\ln^2 x + 2 \ln x)$$ $$f'(x) = 1 + \ln^2 x + 2 \ln x$$ **Step 2: Find the value of $x$ for which $f'(x)$ vanishes.** Set $f'(x) = 0$: $$1 + \ln^2 x + 2 \ln x = 0$$ This is a quadratic equation in terms of $\ln x$. Let $u = \ln x$: $$u^2 + 2u + 1 = 0$$ This is a perfect square trinomial: $$(u + 1)^2 = 0$$ Solving for $u$: $$u = -1$$ **Step 3: Solve for $x$.** Substitute back $u = \ln x$: $$\ln x = -1$$ Exponentiate both sides with base $e$: $$x = e^{-1}$$ $$x = \frac{1}{e}$$ The value of $x$ where $f'(x)$ vanishes is $\frac{1}{e}$.