91Ó°ÊÓ

To find the derivative of \(f(x)\), we use the basic rules of differentiation: - The derivative of a constant is zero. - The derivative of \(x\) is 1. - The derivative of an integral function is the integrand evaluated at \(x\). Differentiating \(f(x) = 1 + x + \int_{1}^{x} (\ln t)^2 + 2 \ln t dt\), we get: \(f'(x) = 0 + 1 + \left((\ln x)^2 + 2 \ln x\right)\)

Our goal is to find the value of the function \(f(x)\) where the derivative \(f'(x)\) vanishes. So we set the derivative of the function equal to zero: \(0 = 1 + (\ln x)^2 + 2 \ln x\)

To find the value of \(x\) where the derivative vanishes, we can reorganize the equation: \((\ln x)^2 + 2 \ln x + 1 = 0\) This is a quadratic equation in terms of \(\ln x\). Let \(u = \ln x\). The equation can be written as: \(u^2 + 2u + 1 = 0\) This equation can be factored as: \((u + 1)^2 = 0\) Therefore, \(u + 1 = 0\), implying that \(u = -1\). Since \(u = \ln x\), we have: \(\ln x = -1\) Taking the exponential of both sides: \(x = e^{-1}\)

Finally, we need to calculate the value of \(f(x)\) when \(x = e^{-1}\). So we plug in the value of \(x\) into the function: \(f(e^{-1}) = 1 + e^{-1} + \int_{1}^{e^{-1}} (\ln t)^2 + 2 \ln t dt\) We can use the properties of logarithms to simplify the integral: \(f(e^{-1}) = 1 + e^{-1} + \int_{1}^{e^{-1}} t^{\ln t} dt\) At this point, unfortunately, there isn't a closed-form antiderivative for this integral. Therefore, we can only approximate the value of the integral using numerical methods such as Riemann sums, trapezoidal rule, or Simpson's rule. Regardless, we have found a general expression for the value of the function where its derivative vanishes: \(f(e^{-1}) \approx 1 + e^{-1} + \int_{1}^{e^{-1}} t^{\ln t} dt\)