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

We are given the functional equation

$$f(x)=\int_{1}^{x} f(t)\,dt+\bigl(1-x\bigr)\bigl(\ln x-1\bigr)+e,\qquad x\ge 1.$$

Put $$F(x)=\int_{1}^{x} f(t)\,dt.$$
Then $$F'(x)=f(x).$$ Substituting this in the given relation,

$$f(x)=F(x)+(1-x)(\ln x-1)+e.\quad -(1)$$

Differentiate $$(1)$$ with respect to $$x$$:

$$f'(x)=F'(x)+\frac{d}{dx}\!\left[(1-x)(\ln x-1)\right].$$

Because $$F'(x)=f(x),$$ the left-hand term becomes $$f(x).$$ Compute the derivative of the remaining part:

$$\frac{d}{dx}\!\left[(1-x)(\ln x-1)\right]=-(\ln x-1)+\frac{1-x}{x} = -\ln x +1 +\frac1x -1 = -\ln x+\frac1x.$$

Hence

$$f'(x)=f(x)-\ln x+\frac1x\qquad\Longrightarrow\qquad f'(x)-f(x)=-\ln x+\frac1x.\quad -(2)$$

Equation $$(2)$$ is a first-order linear ODE of the form $$y'-y=g(x).$$
Its integrating factor is $$e^{-x}.$$ Multiplying throughout by this factor:

$$e^{-x}f'(x)-e^{-x}f(x)=e^{-x}\!\left(-\ln x+\frac1x\right) \quad\Longrightarrow\quad \frac{d}{dx}\!\bigl[e^{-x}f(x)\bigr]=e^{-x}\!\left(-\ln x+\frac1x\right).$$

Integrate from $$1$$ to $$x$$ (where $$x\ge1$$):

$$e^{-x}f(x)-e^{-1}f(1)=\int_{1}^{x}e^{-s}\!\left(-\ln s+\frac1s\right)\,ds.\quad -(3)$$

First evaluate $$f(1)$$ directly from the original relation at $$x=1$$:

$$f(1)=\int_{1}^{1}f(t)\,dt+(1-1)(\ln1-1)+e=0+0+e=e.$$

Using $$f(1)=e$$ in $$(3)$$ gives

$$e^{-x}f(x)=1+\int_{1}^{x}e^{-s}\!\left(-\ln s+\frac1s\right)\,ds.\quad -(4)$$

Observe that

$$\frac{d}{ds}\bigl[e^{-s}\ln s\bigr]=e^{-s}\!\bigl(-\ln s+\frac1s\bigr).$$

Therefore the integral in $$(4)$$ is simply a difference of values:

$$\int_{1}^{x}e^{-s}\!\left(-\ln s+\frac1s\right)\,ds =\Bigl[e^{-s}\ln s\Bigr]_{1}^{x} =e^{-x}\ln x-e^{-1}\ln1=e^{-x}\ln x.$$

Substituting back into $$(4)$$:

$$e^{-x}f(x)=1+e^{-x}\ln x \quad\Longrightarrow\quad f(x)=e^{x}\Bigl(1+e^{-x}\ln x\Bigr)=e^{x}+e^{x}\!\cdot e^{-x}\ln x =e^{x}+\ln x.$$

In particular, for $$x=e$$,

$$f(e)=e^{e}+\ln e=e^{e}+1.$$

But $$f(1)=e,$$ so $$f\bigl(f(1)\bigr)=f(e)=e^{e}+1.$$

Hence $$f(f(1))=1+e^{e}.$$ Therefore the correct choice is:

Option A which is: $$1+e^{e}.$$