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Step-by-Step Solution
Step 1: Express the given function F(x)
We have
$F(x) = f(x) + f\!\bigl(\tfrac{1}{x}\bigr)$
where
$f(x) = \int_{1}^{x} \frac{\log t}{1+t} \,dt.$
We need to find $F(e)$.
Step 2: Write out F(e) explicitly
Substitute $x = e$ into $F(x)$:
$F(e) = f(e) + f\!\bigl(\tfrac{1}{e}\bigr).$
So,
$F(e) = \int_{1}^{e} \frac{\log t}{1 + t} \,dt \;+\; \int_{1}^{\tfrac{1}{e}} \frac{\log t}{1 + t} \,dt.$
Step 3: Transform the second integral
Consider
$I = \int_{1}^{\tfrac{1}{e}} \frac{\log t}{1 + t} \,dt.$
Use the substitution
$z = \frac{1}{t}.$
Then
$dz = -\frac{1}{t^2}\,dt \implies dt = -\frac{dz}{z^2}.$
When $t=1,\, z=1,$ and when $t=\tfrac{1}{e},\, z=e.$
Thus,
$I = \int_{t=1}^{t=\tfrac{1}{e}} \frac{\log t}{1+t}\,dt
= \int_{z=1}^{z=e} \frac{\log\!\bigl(\tfrac{1}{z}\bigr)}{1 + \tfrac{1}{z}}
\Bigl(-\frac{dz}{z^2}\Bigr).$
Since $\log\!\bigl(\tfrac{1}{z}\bigr) = -\log z,$ and $1 + \tfrac{1}{z} = \tfrac{z+1}{z},$ one simplifies carefully to get
$I = \int_{1}^{e} \frac{\log z}{z\,(z+1)}\,dz.
Step 4: Combine the two integrals
Hence,
$F(e) = \underbrace{\int_{1}^{e} \frac{\log t}{1 + t}\,dt}_{\text{first integral}}
\;+\;
\underbrace{\int_{1}^{e} \frac{\log t}{t\,(1 + t)}\,dt}_{\text{transformed second integral}}.
Factor $\log t$ out:
$F(e) = \int_{1}^{e} \log t \Bigl[\frac{1}{1+t} + \frac{1}{t(1+t)}\Bigr]\,dt
= \int_{1}^{e} \frac{\log t\,(t + 1)}{t\,(t+1)}\,dt
= \int_{1}^{e} \frac{\log t}{t}\,dt.$
Step 5: Evaluate the resulting integral
Compute
$\int_{1}^{e} \frac{\log t}{t}\,dt.$
Use the substitution
$x = \log t.$
Then
$dx = \frac{1}{t}\,dt,$
so
$\frac{1}{t}\,dt = dx.$
When $t=1,\, x=0,$ and when $t=e,\, x=1.$
Thus,
$F(e) = \int_{0}^{1} x \,dx = \Bigl[\frac{x^2}{2}\Bigr]_{0}^{1} = \frac{1}{2}.$
Final Answer
$\displaystyle F(e) = \frac{1}{2}.$
