If F(x) = f(x) + fleft(frac{1}{x}right),where | vedclass

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(C) Given $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$. Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$. Let $t = \

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$0.5$

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(C) Given $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$. Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$. Let $t = \frac{1}{u}$,then $dt = -\frac{1}{u^2} du$. When $t=1, u=1$. When $t=1/x, u=x$. $f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{\frac{u+1}{u}} \left(\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{u(u+1)} du$. Now,$F(x) = f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt - \int_{1}^{x} \frac{\log_{e} t}{t(1+t)} dt$. $F(x) = \int_{1}^{x} \log_{e} t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{x} \log_{e} t \left( \frac{t-1}{t(1+t)} \right) dt$. Wait,let's re-evaluate: $F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( 1 - \frac{1}{t} \right) dt = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( \frac{t-1}{t} \right) dt$. Actually,$f(x) + f(1/x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt + \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$. Using the property $\int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} (-du/u^2) = \int_{1}^{x} \frac{\log_{e} u}{u(u+1)} du$. $F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} (1 + 1/t) dt = \int_{1}^{x} \frac{\log_{e} t}{t} dt$. $F(x) = \left[ \frac{(\log_{e} t)^2}{2} \right]_{1}^{x} = \frac{(\log_{e} x)^2}{2}$. For $x=e$,$F(e) = \frac{(\log_{e} e)^2}{2} = \frac{1}{2} = 0.5$.

If $F(x) = f(x) + f\left(\frac{1}{x}\right)$,where $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$,then $F(e) = $

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