Let f: [1, infty) to mathbb{R} be a different | vedclass

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(A) Given the equation $f(x) = \int_1^x f(t) \, dt + (1 - x)(\log_e x - 1) + e$.At $x = 1$, $f(1) = \int_1^1 f(t) \, dt + (1 - 1)(\log_e 1 - 1) + e =

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$1 + e^e$

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(A) Given the equation $f(x) = \int_1^x f(t) \, dt + (1 - x)(\log_e x - 1) + e$.At $x = 1$, $f(1) = \int_1^1 f(t) \, dt + (1 - 1)(\log_e 1 - 1) + e = 0 + 0 + e = e$.Differentiating both sides with respect to $x$ using the Leibniz rule:$f'(x) = f(x) - 1(\log_e x - 1) + (1 - x)(1/x) = f(x) - \log_e x + 1 + 1/x - 1 = f(x) - \log_e x + 1/x$.Rearranging gives the linear differential equation: $f'(x) - f(x) = \frac{1}{x} - \log_e x$.The integrating factor is $I.F. = e^{\int -1 \, dx} = e^{-x}$.Multiplying by $e^{-x}$: $\frac{d}{dx}(f(x)e^{-x}) = e^{-x}(\frac{1}{x} - \log_e x)$.Integrating both sides: $f(x)e^{-x} = \int e^{-x}(\frac{1}{x} - \log_e x) \, dx + C$.Using the property $\frac{d}{dx}(e^{-x} \log_e x) = -e^{-x} \log_e x + e^{-x}/x$, we see that $f(x)e^{-x} = e^{-x} \log_e x + C$.Thus, $f(x) = \log_e x + Ce^x$.Since $f(1) = e$, we have $e = \log_e 1 + Ce^1 \implies e = 0 + Ce \implies C = 1$.So, $f(x) = \log_e x + e^x$.We need to find $f(f(1)) = f(e) = \log_e e + e^e = 1 + e^e$.

Let $f: [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t) \, dt + (1 - x)(\log_e x - 1) + e$. Then the value of $f(f(1))$ is:

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