For $$x > 0$$, if $$f(x) = \int_1^x \frac{\log_e t}{(1+t)} dt$$, then $$f(e) + f\left(\frac{1}{e}\right)$$ is equal to:

We have $$f(x) = \displaystyle\int_1^x \dfrac{\ln t}{1 + t}\,dt$$ for $$x > 0$$.

To find $$f\left(\dfrac{1}{e}\right)$$, substitute $$t = \dfrac{1}{u}$$ so that $$dt = -\dfrac{du}{u^2}$$ and $$\ln t = -\ln u$$. When $$t = 1$$, $$u = 1$$; when $$t = \dfrac{1}{e}$$, $$u = e$$.

$$f\left(\dfrac{1}{e}\right) = \displaystyle\int_1^{1/e} \dfrac{\ln t}{1+t}\,dt = \displaystyle\int_1^{e} \dfrac{-\ln u}{1 + \frac{1}{u}} \cdot \left(-\dfrac{du}{u^2}\right) = \displaystyle\int_1^{e} \dfrac{\ln u}{u^2 \cdot \frac{u+1}{u}}\,du = \displaystyle\int_1^{e} \dfrac{\ln u}{u(u+1)}\,du$$.

Now, $$f(e) + f\left(\dfrac{1}{e}\right) = \displaystyle\int_1^e \dfrac{\ln t}{1+t}\,dt + \displaystyle\int_1^e \dfrac{\ln t}{t(t+1)}\,dt$$.

Combining the integrands: $$\dfrac{\ln t}{1+t} + \dfrac{\ln t}{t(t+1)} = \ln t \cdot \dfrac{t + 1}{t(t+1)} = \dfrac{\ln t}{t}$$.

Therefore $$f(e) + f\left(\dfrac{1}{e}\right) = \displaystyle\int_1^e \dfrac{\ln t}{t}\,dt = \left[\dfrac{(\ln t)^2}{2}\right]_1^e = \dfrac{1}{2} - 0 = \dfrac{1}{2}$$.