The value of the function f(x) = 1 + x + intl | vedclass

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(D) Given $f(x) = 1 + x + \int\limits_1^x (\ln^2 t + 2 \ln t) \, dt$. Applying Leibniz's rule to differentiate with respect to $x$: $f'(x) = 0 + 1 + (

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$1 + 2 e^{-1}$

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(D) Given $f(x) = 1 + x + \int\limits_1^x (\ln^2 t + 2 \ln t) \, dt$. Applying Leibniz's rule to differentiate with respect to $x$: $f'(x) = 0 + 1 + (\ln^2 x + 2 \ln x) = 0$. This simplifies to $(\ln x + 1)^2 = 0$,which gives $\ln x = -1$,so $x = e^{-1}$. Now,evaluate $f(e^{-1}) = 1 + e^{-1} + \int\limits_1^{e^{-1}} (\ln^2 t + 2 \ln t) \, dt$. Note that $\frac{d}{dt}(t \ln^2 t) = \ln^2 t + t(2 \ln t \cdot \frac{1}{t}) = \ln^2 t + 2 \ln t$. Thus,$\int (\ln^2 t + 2 \ln t) \, dt = t \ln^2 t + C$. Evaluating the definite integral: $f(e^{-1}) = 1 + e^{-1} + [t \ln^2 t]_1^{e^{-1}} = 1 + e^{-1} + (e^{-1} \cdot (-1)^2 - 1 \cdot 0^2) = 1 + e^{-1} + e^{-1} = 1 + 2e^{-1}$.

The value of the function $f(x) = 1 + x + \int\limits_1^x (\ln^2 t + 2 \ln t) \, dt$ where $f'(x)$ vanishes is:

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