Let f: [1, ∞) arrow R be a differentiable function defined as f(x)...

Given $f(x) = \int_1^x f(t)\,dt + (1-x)(\log_e x - 1) + e$

Substituting $x = 1$ in the given equation:

$f(1) = \int_1^1 f(t)\,dt + (1-1)(\log_e 1 - 1) + e$

$f(1) = 0 + 0 + e = e$

Differentiating the given equation with respect to $x$ using the Leibniz rule:

$f'(x) = f(x) + \dfrac{d}{dx}[(1-x)(\log_e x - 1)]$

$f'(x) = f(x) + (-1)(\log_e x - 1) + (1-x)\left(\dfrac{1}{x}\right)$

$f'(x) = f(x) - \log_e x + 1 + \dfrac{1}{x} - 1$

$f'(x) - f(x) = \dfrac{1}{x} - \log_e x$

This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = -1$ and $Q = \dfrac{1}{x} - \log_e x$.

Integrating Factor (IF) $= e^{\int -1 \,dx} = e^{-x}$

Multiplying the differential equation by the IF:

$e^{-x} f'(x) - e^{-x} f(x) = e^{-x} \left(\dfrac{1}{x} - \log_e x\right)$

$\dfrac{d}{dx}[e^{-x} f(x)] = e^{-x} \dfrac{1}{x} - e^{-x} \log_e x$

Notice that $\dfrac{d}{dx}[e^{-x} \log_e x] = -e^{-x} \log_e x + e^{-x} \dfrac{1}{x}$.

Thus, $\dfrac{d}{dx}[e^{-x} f(x)] = \dfrac{d}{dx}[e^{-x} \log_e x]$

Integrating both sides with respect to $x$:

$e^{-x} f(x) = e^{-x} \log_e x + C$

$f(x) = \log_e x + C e^x$

Using the initial condition $f(1) = e$:

$e = \log_e 1 + C e^1$

$e = 0 + C e \Rightarrow C = 1$

Therefore, the function is $f(x) = \log_e x + e^x$.

We need to find $f(f(1))$:

Since $f(1) = e$, we evaluate $f(e)$:

$f(e) = \log_e e + e^e = 1 + e^e$

Answer: $(1 + e^e)$